NEWTONIAN GRAVITY IN THE LANGUAGE OF CURVED SPACETIME

The longest period of time for which a modern painting has hungupside down in a public gallery unnoticed is 47 days. This occurred to Le Bateau by Matisse in the Museum of Modern Art, New York City. In this time 116,000 people had passed through the gallery.McWHIRTER AND McWHIRTER (1971)

§12.1. NEWTONIAN GRAVITY IN BRIEF

The equivalence principle is not unique to Einstein's description of the facts of gravity. What is unique to Einstein is the combination of the equivalence principle and local Lorentz geometry. To return to the world of Newton, forget everything discovered in the last century about special relativity, light cones, the limiting speed of light, and proper time. Return to the "universal time" t t ttt of earlier centuries. In terms of that universal time, and of rectangular, "Galilean" space coordinates, Newtonian theory gives for the trajectories of neutral test particles
(12.1) d 2 x j d t 2 + Φ x j = 0 (12.1) d 2 x j d t 2 + Φ x j = 0 {:(12.1)(d^(2)x^(j))/(dt^(2))+(del Phi)/(delx^(j))=0:}\begin{equation*} \frac{d^{2} x^{j}}{d t^{2}}+\frac{\partial \Phi}{\partial x^{j}}=0 \tag{12.1} \end{equation*}(12.1)d2xjdt2+Φxj=0
Φ ( Φ ( Phi(\Phi(Φ( sometimes denoted U ) = U ) = -U)=-U)=U)= Newtonian potential.
Customarily one interprets these equations as describing the "curved paths" x j ( t ) x j ( t ) x^(j)(t)x^{j}(t)xj(t) along which test particles fall in Euclidean space (not spacetime). These curved paths include circular orbits about the Earth and the parabolic trajectory of a baseball. Cartan ( 1923 , 1924 ) ( 1923 , 1924 ) (1923,1924)(1923,1924)(1923,1924) asks one to abandon this viewpoint. Instead, he says, regard these trajectories as geodesics [ t ( λ ) , x j ( λ ) ] t ( λ ) , x j ( λ ) [t(lambda),x^(j)(lambda)]\left[t(\lambda), x^{j}(\lambda)\right][t(λ),xj(λ)] in curved spacetime. (This change of viewpoint was embodied in Figures B and C of Box 1.6.) Since the "affinely ticking"
This chapter is entirely Track 2. Chapters 9-11 are necessary preparation for it. It is not needed for any later chapter, but it will be helpful in
(1) Chapter 17 (Einstein field equations) and
(2) Chapters 38 and 39 (experimental tests and other theories of gravity).
Newtonian gravity: original formulation
Newtonian gravity: translation into language of curved spacetime
Newtonian clocks carried by test particles read universal time (or some multiple, λ = a t + b λ = a t + b lambda=at+b\lambda=a t+bλ=at+b, thereof), the equation of motion (12.1) can be rewritten
(12.3) d 2 t d λ 2 = 0 , d 2 x j d λ 2 + Φ x j ( d t d λ ) 2 = 0 (12.3) d 2 t d λ 2 = 0 , d 2 x j d λ 2 + Φ x j d t d λ 2 = 0 {:(12.3)(d^(2)t)/(dlambda^(2))=0","quad(d^(2)x^(j))/(dlambda^(2))+(del Phi)/(delx^(j))((dt)/(d lambda))^(2)=0:}\begin{equation*} \frac{d^{2} t}{d \lambda^{2}}=0, \quad \frac{d^{2} x^{j}}{d \lambda^{2}}+\frac{\partial \Phi}{\partial x^{j}}\left(\frac{d t}{d \lambda}\right)^{2}=0 \tag{12.3} \end{equation*}(12.3)d2tdλ2=0,d2xjdλ2+Φxj(dtdλ)2=0
By comparing with the geodesic equation
d 2 x α / d λ 2 + Γ α β γ ( d x β / d λ ) ( d x γ / d λ ) = 0 , d 2 x α / d λ 2 + Γ α β γ d x β / d λ d x γ / d λ = 0 , d^(2)x^(alpha)//dlambda^(2)+Gamma^(alpha)_(beta gamma)(dx^(beta)//d lambda)(dx^(gamma)//d lambda)=0,d^{2} x^{\alpha} / d \lambda^{2}+\Gamma^{\alpha}{ }_{\beta \gamma}\left(d x^{\beta} / d \lambda\right)\left(d x^{\gamma} / d \lambda\right)=0,d2xα/dλ2+Γαβγ(dxβ/dλ)(dxγ/dλ)=0,
one can read off the values of the connection coefficients:
(12.4) Γ j 00 = Φ / x j ; all other Γ β γ α vanish. (12.4) Γ j 00 = Φ / x j ;  all other  Γ β γ α  vanish.  {:(12.4)Gamma^(j)_(00)=del Phi//delx^(j);quad" all other "Gamma_(beta gamma)^(alpha)" vanish. ":}\begin{equation*} \Gamma^{j}{ }_{00}=\partial \Phi / \partial x^{j} ; \quad \text { all other } \Gamma_{\beta \gamma}^{\alpha} \text { vanish. } \tag{12.4} \end{equation*}(12.4)Γj00=Φ/xj; all other Γβγα vanish. 
And by inserting these into the standard equation (11.12) for the components of the Riemann tensor, one learns (exercise 12.1)
(12.5) R j 0 k 0 = R j 00 k = 2 Φ x j x k ; all other R α β γ δ vanish. (12.5) R j 0 k 0 = R j 00 k = 2 Φ x j x k ;  all other  R α β γ δ  vanish.  {:(12.5)R^(j)_(0k0)=-R^(j)_(00 k)=(del^(2)Phi)/(delx^(j)delx^(k));quad" all other "R^(alpha)_(beta gamma delta)" vanish. ":}\begin{equation*} R^{j}{ }_{0 k 0}=-R^{j}{ }_{00 k}=\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{k}} ; \quad \text { all other } R^{\alpha}{ }_{\beta \gamma \delta} \text { vanish. } \tag{12.5} \end{equation*}(12.5)Rj0k0=Rj00k=2Φxjxk; all other Rαβγδ vanish. 
Finally, the source equation for the Newtonian potential
(12.6) 2 Φ j Φ , j j = 4 π ρ (12.6) 2 Φ j Φ , j j = 4 π ρ {:(12.6)grad^(2)Phi-=sum_(j)Phi_(,jj)=4pi rho:}\begin{equation*} \nabla^{2} \Phi \equiv \sum_{j} \Phi_{, j j}=4 \pi \rho \tag{12.6} \end{equation*}(12.6)2ΦjΦ,jj=4πρ
one can rewrite with the help of the "Ricci curvature tensor"
(12.7) R α β R μ α μ β ( contraction of Riemann ) (12.7) R α β R μ α μ β (  contraction of Riemann  ) {:(12.7)R_(alpha beta)-=R^(mu)_(alpha mu beta)(" contraction of Riemann "):}\begin{equation*} R_{\alpha \beta} \equiv R^{\mu}{ }_{\alpha \mu \beta}(\text { contraction of Riemann }) \tag{12.7} \end{equation*}(12.7)RαβRμαμβ( contraction of Riemann )
in the geometric form (exercise 12.2)
(12.8) R 00 = 4 π ρ ; all other R α β vanish. (12.8) R 00 = 4 π ρ ;  all other  R α β  vanish.  {:(12.8)R_(00)=4pi rho;quad" all other "R_(alpha beta)" vanish. ":}\begin{equation*} R_{00}=4 \pi \rho ; \quad \text { all other } R_{\alpha \beta} \text { vanish. } \tag{12.8} \end{equation*}(12.8)R00=4πρ; all other Rαβ vanish. 
Equation (12.4) for Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ, equation (12.5) for R α β γ δ R α β γ δ R^(alpha)_(beta gamma delta)R^{\alpha}{ }_{\beta \gamma \delta}Rαβγδ, equation (12.8) for R α β R α β R_(alpha beta)R_{\alpha \beta}Rαβ, plus the law of geodesic motion are the full content of Newtonian gravity, rewritten in geometric language.
It is one thing to pass quickly through these component manipulations. It is quite another to understand fully, in abstract and pictorial terms, the meanings of these equations and the structure of Newtonian spacetime. To produce such understanding, and to compare Newtonian spacetime with Einsteinian spacetime, are the goals of this chapter, which is based on the work of Cartan (1923, 1924), Trautman (1965), and Misner (1969a).

EXERCISES

Exercise 12.1. RIEMANN CURVATURE OF NEWTONIAN SPACETIME

Derive equation (12.5) for R α β γ δ R α β γ δ R^(alpha)_(beta gamma delta)R^{\alpha}{ }_{\beta \gamma \delta}Rαβγδ from equation (12.4) for Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ.

Exercise 12.2. NEWTONIAN FIELD EQUATION

Derive the geometric form (12.8) of the Newtonian field equation from (12.5) through (12.7).

§12.2. STRATIFICATION OF NEWTONIAN SPACETIME

Galileo and Newton spoke of a flat, Euclidean "absolute space" and of an "absolute time," two concepts distinct and unlinked. In absolute space Newtonian physics took place; and as it took place, absolute time marched on. No hint was there that space and time might be two aspects of a single entity, a curved "spacetime"-until Einstein made the unification in relativity physics, and Cartan (1923) followed suit in Newtonian physics in order to provide clearer insight into Einstein's ideas.
How do the absolute space of Galileo and Newton, and their absolute time, fit into Cartan's "Newtonian spacetime"? The key to the fit is stratification; stratification produced by the universal time coordinate t t ttt.
Regard t t ttt as a function (scalar field) defined once and for all in Newtonian spacetime
(12.9) t = t ( P ) (12.9) t = t ( P ) {:(12.9)t=t(P):}\begin{equation*} t=t(\mathscr{P}) \tag{12.9} \end{equation*}(12.9)t=t(P)
Without it, spacetime could not be Newtonian, for " t t ttt " is every bit as intrinsic to Newtonian spacetime as the metric " g g g\boldsymbol{g}g " is to Lorentz spacetime. The layers of spacetime are the slices of constant t t ttt-the "space slices"-each of which has an identical geometric structure: the old "absolute space."
Adopting Cartan's viewpoint, ask what kind of geometry is induced onto each space slice by the surrounding geometry of spacetime. A given space slice is endowed, by the Galilean coordinates of § 12.1 § 12.1 §12.1\S 12.1§12.1, with basis vectors e j = / x j e j = / x j e_(j)=del//delx^(j)\boldsymbol{e}_{j}=\partial / \partial x^{j}ej=/xj; and this basis has vanishing connection coefficients, Γ j k = 0 Γ j k = 0 Gamma^(j)_(kℓ)=0\Gamma^{j}{ }_{k \ell}=0Γjk=0 [cf. equation (12.4)]. Consequently, the geometry of each space slice is completely flat.
"Absolute space" is Euclidean in its geometry, according to the old viewpoint, and the Galilean coordinates are Cartesian. Translated into Cartan's language, this says: not only is each space slice ( t = t = t=t=t= constant) flat, and not only do its Galilean coordinates have vanishing connection coefficients, but also each space slice is endowed with a three-dimensional metric, and its Galilean coordinate basis is orthonormal,
(12.10) e i e j = ( / x i ) ( / x j ) = δ i j (12.10) e i e j = / x i / x j = δ i j {:(12.10)e_(i)*e_(j)=(del//delx^(i))*(del//delx^(j))=delta_(ij):}\begin{equation*} \boldsymbol{e}_{i} \cdot \boldsymbol{e}_{j}=\left(\partial / \partial x^{i}\right) \cdot\left(\partial / \partial x^{j}\right)=\delta_{i j} \tag{12.10} \end{equation*}(12.10)eiej=(/xi)(/xj)=δij
If the space slices are really so flat, where do curvature and geodesic deviation enter in? They are properties of spacetime. Parallel transport a vector around a closed curve lying entirely in a space slice; it will return to its starting point unchanged. But transport it forward in time by Δ t Δ t Delta t\Delta tΔt, northerly in space by Δ x k Δ x k Deltax^(k)\Delta x^{k}Δxk, back in time by Δ t Δ t -Delta t-\Delta tΔt, and southerly by Δ x k Δ x k -Deltax^(k)-\Delta x^{k}Δxk to its starting point; it will return changed by
δ A = ( Δ t t , Δ x k x k ) A ; δ A = Δ t t , Δ x k x k A ; delta A=-ℜ(Delta t(del)/(del t),Deltax^(k)(del)/(delx^(k)))A;\delta \boldsymbol{A}=-\Re\left(\Delta t \frac{\partial}{\partial t}, \Delta x^{k} \frac{\partial}{\partial x^{k}}\right) \boldsymbol{A} ;δA=(Δtt,Δxkxk)A;
i.e.,
(12.11) δ A 0 = 0 , δ A j = R j 00 k A 0 ( Δ t ) ( Δ x k ) = 2 Φ x j x k A 0 ( Δ t ) ( Δ x k ) (12.11) δ A 0 = 0 , δ A j = R j 00 k A 0 ( Δ t ) Δ x k = 2 Φ x j x k A 0 ( Δ t ) Δ x k {:(12.11)deltaA^(0)=0","quad deltaA^(j)=-R^(j)_(00 k)A^(0)(Delta t)(Deltax^(k))=(del^(2)Phi)/(delx^(j)delx^(k))A^(0)(Delta t)(Deltax^(k)):}\begin{equation*} \delta A^{0}=0, \quad \delta A^{j}=-R^{j}{ }_{00 k} A^{0}(\Delta t)\left(\Delta x^{k}\right)=\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{k}} A^{0}(\Delta t)\left(\Delta x^{k}\right) \tag{12.11} \end{equation*}(12.11)δA0=0,δAj=Rj00kA0(Δt)(Δxk)=2ΦxjxkA0(Δt)(Δxk)
Geodesics of a space slice (Euclid's straight lines) that are initially parallel remain
The geometry of Newtonian spacetime:
"Universal time" as a scalar field
Space slices with Euclidean geometry
Curvature acts in spacetime, not in space slices
always parallel. But geodesics of spacetime (trajectories of freely falling particles) initially parallel get pried apart or pushed together by spacetime curvature,
u u n + ( n , u ) u = 0 u u n + ( n , u ) u = 0 grad_(u)grad_(u)n+ℜ(n,u)u=0\boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{u} \boldsymbol{n}+\Re(\boldsymbol{n}, \boldsymbol{u}) \boldsymbol{u}=0uun+(n,u)u=0
or equivalently in Galilean coordinates:
(12.12a) n 0 = d n 0 / d t = 0 initially n 0 = 0 always; (12.12b) d 2 n j d t 2 + 2 Φ x j x k n k = 0 (12.12a) n 0 = d n 0 / d t = 0  initially  n 0 = 0  always;  (12.12b) d 2 n j d t 2 + 2 Φ x j x k n k = 0 {:[(12.12a)n^(0)=dn^(0)//dt=0" initially "Longrightarrown^(0)=0" always; "],[(12.12b)(d^(2)n^(j))/(dt^(2))+(del^(2)Phi)/(delx^(j)delx^(k))n^(k)=0]:}\begin{align*} n^{0}=d n^{0} / d t=0 \text { initially } \Longrightarrow & n^{0}=0 \text { always; } \tag{12.12a}\\ & \frac{d^{2} n^{j}}{d t^{2}}+\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{k}} n^{k}=0 \tag{12.12b} \end{align*}(12.12a)n0=dn0/dt=0 initially n0=0 always; (12.12b)d2njdt2+2Φxjxknk=0
(see Box 12.1 and exercise 12.3).

EXERCISE

Exercise 12.3. GEODESIC DEVIATION DERIVED

Produce a third column for Box 11.4, one that carries out the "geometric analysis" in component notation using the Galilean connection coefficients (12.4) of Newtonian spacetime. Thereby achieve a deeper understanding of how the geometric analysis parallels the old Newtonian analysis.
Galilean coordinates defined

§12.3. GALILEAN COORDINATE SYSTEMS

The Lorentz spacetime of special relativity has an existence and structure completely independent of any coordinate system. But a special property of its geometry (zero curvature) allows the introduction of a special class of coordinates (Lorentz coordinates), which cling to spacetime in a special way
( / x α ) ( / x β ) = η α β everywhere. / x α / x β = η α β  everywhere.  (del//delx^(alpha))*(del//delx^(beta))=eta_(alpha beta)" everywhere. "\left(\partial / \partial x^{\alpha}\right) \cdot\left(\partial / \partial x^{\beta}\right)=\eta_{\alpha \beta} \text { everywhere. }(/xα)(/xβ)=ηαβ everywhere. 
By studying these special coordinate systems and the relationships between them (Lorentz transformations), one learns much about the structure of spacetime itself (breakdown in simultaneity; Lorentz contraction; time dilatation; . . .).
Similarly for Newtonian spacetime. Special properties of its geometry (explored in abstract later; Box 12.4) permit the introduction of special coordinates (Galilean coordinates), which cling to spacetime in a special way
x 0 ( P ) = t ( P ) ; ( / x j ) ( / x k ) = δ j k ; x 0 ( P ) = t ( P ) ; / x j / x k = δ j k ; {:[x^(0)(P)=t(P);],[(del//delx^(j))*(del//delx^(k))=delta_(jk);]:}\begin{gathered} x^{0}(\mathscr{P})=t(\mathscr{P}) ; \\ \left(\partial / \partial x^{j}\right) \cdot\left(\partial / \partial x^{k}\right)=\delta_{j k} ; \end{gathered}x0(P)=t(P);(/xj)(/xk)=δjk;
Γ j 00 = Φ , j for some scalar field Φ , and all other Γ α β γ vanish. Γ j 00 = Φ , j  for some scalar field  Φ ,  and all other  Γ α β γ  vanish.  Gamma^(j)_(00)=Phi_(,j)" for some scalar field "Phi," and all other "Gamma^(alpha)_(beta gamma)" vanish. "\Gamma^{j}{ }_{00}=\Phi_{, j} \text { for some scalar field } \Phi, \text { and all other } \Gamma^{\alpha}{ }_{\beta \gamma} \text { vanish. }Γj00=Φ,j for some scalar field Φ, and all other Γαβγ vanish. 
To understand Newtonian spacetime more deeply, study the relations between these Galilean coordinate systems.

Box 12.1 GEODESIC DEVIATION IN NEWTONIAN SPACETIME

Coordinate system for calculation: Galilean space coordinates x j x j x^(j)x^{j}xj and universal time coordinate t t ttt. General component form of equation:
D 2 n α d λ 2 + R α β γ δ d x β d λ n γ d x δ d λ = 0 . D 2 n α d λ 2 + R α β γ δ d x β d λ n γ d x δ d λ = 0 . (D^(2)n^(alpha))/(dlambda^(2))+R^(alpha)_(beta gamma delta)(dx^(beta))/(d lambda)n^(gamma)(dx^(delta))/(d lambda)=0.\frac{D^{2} n^{\alpha}}{d \lambda^{2}}+R^{\alpha}{ }_{\beta \gamma \delta} \frac{d x^{\beta}}{d \lambda} n^{\gamma} \frac{d x^{\delta}}{d \lambda}=0 .D2nαdλ2+Rαβγδdxβdλnγdxδdλ=0.
Special conditions for this calculation: let the particles' clocks (affine parameters) all be normalized to read universal time, λ = t λ = t lambda=t\lambda=tλ=t. This means that the separation vector
n α = ( x α / n ) λ n α = x α / n λ n^(alpha)=(delx^(alpha)//del n)_(lambda)n^{\alpha}=\left(\partial x^{\alpha} / \partial n\right)_{\lambda}nα=(xα/n)λ
between geodesics has zero time component, n 0 = 0 n 0 = 0 n^(0)=0n^{0}=0n0=0; i.e., in abstract language,
d t , n = t , α n α = n 0 = 0 ; d t , n = t , α n α = n 0 = 0 ; (:dt,n:)=t_(,alpha)n^(alpha)=n^(0)=0;\langle\boldsymbol{d} t, \boldsymbol{n}\rangle=t_{, \alpha} n^{\alpha}=n^{0}=0 ;dt,n=t,αnα=n0=0;
i.e., in geometric language, n n n\boldsymbol{n}n lies in a space slice
(surface of constant t t ttt ).
Evaluation of covariant derivative:
Evaluation of tidal accelerations:
R 0 β γ δ d x β d λ n γ d x δ d λ = 0 R 0 β γ δ d x β d λ n γ d x δ d λ = 0 R^(0)_(beta gamma delta)(dx^(beta))/(d lambda)n^(gamma)(dx^(delta))/(d lambda)=0quadR^{0}{ }_{\beta \gamma \delta} \frac{d x^{\beta}}{d \lambda} n^{\gamma} \frac{d x^{\delta}}{d \lambda}=0 \quadR0βγδdxβdλnγdxδdλ=0 since R j 0 k 0 R j 0 k 0 R^(j)_(0k0)R^{j}{ }_{0 k 0}Rj0k0 and R j 00 k R j 00 k R^(j)_(00 k)R^{j}{ }_{00 k}Rj00k are only nonzero components.
R j β γ γ δ d x β d λ n γ d x δ d λ = R j 0 k 0 d t d λ n k d t d λ = R j 0 k 0 n k = 2 Φ x j x k n k . [ 0 unless γ is space index] R j β γ γ δ d x β d λ n γ d x δ d λ = R j 0 k 0 d t d λ n k d t d λ = R j 0 k 0 n k = 2 Φ x j x k n k . [ 0  unless  γ  is space index]  ubrace(R^(j)_(beta gamma gamma delta)(dx^(beta))/(d lambda)n^(gamma)(dx^(delta))/(d lambda)=R^(j)_(0k0)(dt)/(d lambda)n^(k)(dt)/(d lambda)=R^(j)_(0k0)n^(k)=(del^(2)Phi)/(delx^(j)delx^(k))n^(k).ubrace)_(uarr){:[0" unless "gamma" is space index] ":}\underbrace{R^{j}{ }_{\beta \gamma \gamma \delta} \frac{d x^{\beta}}{d \lambda} n^{\gamma} \frac{d x^{\delta}}{d \lambda}=R^{j}{ }_{0 k 0} \frac{d t}{d \lambda} n^{k} \frac{d t}{d \lambda}=R^{j}{ }_{0 k 0} n^{k}=\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{k}} n^{k} .}_{\uparrow} \begin{aligned} & {[0 \text { unless } \gamma \text { is space index] }}\end{aligned}Rjβγγδdxβdλnγdxδdλ=Rj0k0dtdλnkdtdλ=Rj0k0nk=2Φxjxknk.[0 unless γ is space index] 
[for γ γ gamma\gammaγ a space index: 0 unless β = δ = 0 β = δ = 0 beta=delta=0\beta=\delta=0β=δ=0 ]
Resultant equation of geodesic deviation:
d 2 n 0 d t 2 = 0 ( agrees with result n 0 = 0 always, which followed from choice λ = t for all particles ) d 2 n j d t 2 + 2 Φ x j x k n k = 0 ( agrees with Newton-type calculation in Box 11.4; see also exercise 12.3 ) . d 2 n 0 d t 2 = 0 (  agrees with result  n 0 = 0  always, which   followed from choice  λ = t  for all particles  ) d 2 n j d t 2 + 2 Φ x j x k n k = 0 (  agrees with Newton-type calculation   in Box 11.4; see also exercise 12.3  ) . {:[(d^(2)n^(0))/(dt^(2))=0((" agrees with result "n^(0)=0" always, which ")/(" followed from choice "lambda=t" for all particles "))],[(d^(2)n^(j))/(dt^(2))+(del^(2)Phi)/(delx^(j)delx^(k))n^(k)=0((" agrees with Newton-type calculation ")/(" in Box 11.4; see also exercise 12.3 ")).]:}\begin{aligned} \frac{d^{2} n^{0}}{d t^{2}}=0 & \binom{\text { agrees with result } n^{0}=0 \text { always, which }}{\text { followed from choice } \lambda=t \text { for all particles }} \\ \frac{d^{2} n^{j}}{d t^{2}}+\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{k}} n^{k}=0 & \binom{\text { agrees with Newton-type calculation }}{\text { in Box 11.4; see also exercise 12.3 }} . \end{aligned}d2n0dt2=0( agrees with result n0=0 always, which  followed from choice λ=t for all particles )d2njdt2+2Φxjxknk=0( agrees with Newton-type calculation  in Box 11.4; see also exercise 12.3 ).
Point of principle: how can one write down the laws of gravity and properties of spacetime in Galilean coordinates first ($12.1), and only afterward (here) come to grip with the nature of the coordinate system and its nonuniqueness? Answer: (a quotation from § 3.1 § 3.1 §3.1\S 3.1§3.1, slightly modified): "Here and elsewhere in science, as emphasized not least by Henri Poincaré, that view is out of date which used to say 'Define your terms before you proceed.' All the laws and theories of physics, including Newton's laws of gravity, have this deep and subtle character, that they both define the concepts they use (here Galilean coordinates) and make statements about these concepts."
The Newtonian laws of gravity, written in a Galilean coordinate system
x 0 = t , ( / x j ) ( / x k ) = δ j k x 0 = t , / x j / x k = δ j k x^(0)=t,quad(del//delx^(j))*(del//delx^(k))=delta_(jk)x^{0}=t, \quad\left(\partial / \partial x^{j}\right) \cdot\left(\partial / \partial x^{k}\right)=\delta_{j k}x0=t,(/xj)(/xk)=δjk
make the statement " Γ j 00 = Φ , j Γ j 00 = Φ , j Gamma^(j)_(00)=Phi_(,j)\Gamma^{j}{ }_{00}=\Phi_{, j}Γj00=Φ,j and all other Γ α β γ = 0 Γ α β γ = 0 Gamma^(alpha)_(beta gamma)=0\Gamma^{\alpha}{ }_{\beta \gamma}=0Γαβγ=0 " about the geometry of spacetime. This statement in turn gives information about the relationships between different Galilean systems. Let one Galilean system { x α ( P ) } x α ( P ) {x^(alpha)(P)}\left\{x^{\alpha}(\mathscr{P})\right\}{xα(P)} be given, and seek the most general coordinate transformation leading to another, { x α ( P ) } x α ( P ) {x^(alpha^('))(P)}\left\{x^{\alpha^{\prime}}(\mathscr{P})\right\}{xα(P)}. The following constraints exist: (1) x 0 0 = x 0 = t x 0 0 = x 0 = t x^(0^(0))=x^(0)=tx^{0^{0}}=x^{0}=tx00=x0=t (both time coordinates must be universal time); (2) at fixed t t ttt (i.e., in a fixed space slice) both sets of space coordinates must be Euclidean, so they must be related by a rotation and a translation:
(12.13b) x k = A j k x j a k , with a k A j k a j . (12.13b) x k = A j k x j a k , with  a k A j k a j . {:(12.13b)x^(k)=A_(j^(')k)x^(j^('))-a^(k)", with "a^(k)-=A_(j^(')k)a^(j^(')).:}\begin{align*} & x^{k}=A_{j^{\prime} k} x^{j^{\prime}}-a^{k} \text {, with } a^{k} \equiv A_{j^{\prime} k} a^{j^{\prime}} . \tag{12.13b} \end{align*}(12.13b)xk=Ajkxjak, with akAjkaj.
The rotation and translation might, a priori, be different on different slices, A j k = A j k = A_(j^(')k)=A_{j^{\prime} k}=Ajk= A j k ( t ) A j k ( t ) A_(j^(')k)(t)A_{j^{\prime} k}(t)Ajk(t) and a j = a j ( t ) a j = a j ( t ) a^(j)=a^(j)(t)a^{j}=a^{j}(t)aj=aj(t); but (3) they must be constrained by the required special form of the connection coefficients. Calculate the connection coefficients in the new coordinate system, given their form in the old. The result (exercise 12.4) is:
Γ j 0 k = Γ j k 0 = A j A ˙ k (produces "Coriolis forces"); Γ j 0 0 = Φ x j + A j k ( A ¨ k k x x a ¨ k ) ; ["centrifugal forces"] all other Γ α β γ vanish ""inertial forces"] all Γ j 0 k = Γ j k 0 = A j A ˙ k  (produces "Coriolis forces");  Γ j 0 0 = Φ x j + A j k A ¨ k k x x a ¨ k ;  ["centrifugal forces"]  all other  Γ α β γ  vanish   ""inertial forces"]   all  {:[Gamma^(j^('))_(0^(')k^('))=Gamma^(j^('))_(k^(')0^('))=A_(j^(')ℓ)A^(˙)_(k^(')ℓ)quad" (produces "Coriolis forces"); "],[Gamma^(j^(')_(0)^(')0^('))=(del Phi)/(delx^(j^(')))+A_(j^(')k)(A^(¨)_(k^(')k^('))x^(x^('))-a^(¨)^(k));],[" ["centrifugal forces"] "^("all other "Gamma^(alpha^('))_(beta^(')gamma^('))" vanish "){:" ""inertial forces"] ":}],[" all "]:}\begin{aligned} & \Gamma^{j^{\prime}}{ }_{0^{\prime} k^{\prime}}=\Gamma^{j^{\prime}}{ }_{k^{\prime} 0^{\prime}}=A_{j^{\prime} \ell} \dot{A}_{k^{\prime} \ell} \quad \text { (produces "Coriolis forces"); } \\ & \Gamma^{j^{\prime}{ }_{0}{ }^{\prime} 0^{\prime}}=\frac{\partial \Phi}{\partial x^{j^{\prime}}}+A_{j^{\prime} k}\left(\ddot{A}_{k^{\prime} k^{\prime}} x^{x^{\prime}}-\ddot{a}^{k}\right) ; \\ & \text { ["centrifugal forces"] }{ }^{\text {all other } \Gamma^{\alpha^{\prime}}{ }_{\beta^{\prime} \gamma^{\prime}} \text { vanish }} \begin{array}{l} \text { ""inertial forces"] } \end{array} \\ & \text { all } \end{aligned}Γj0k=Γjk0=AjA˙k (produces "Coriolis forces"); Γj00=Φxj+Ajk(A¨kkxxa¨k); ["centrifugal forces"] all other Γαβγ vanish  ""inertial forces"]  all 
("Euclidean" index conventions; repeated space indices to be summed even if both are down; dot denotes time derivative). These have the standard Galilean form (12.4) if and only if
(12.15) A ˙ j k = 0 , ϕ = Φ a ¨ k x k + constant. [ Newtonian potential in new coordinate system ] [ Newtonian potential in old coordinate system ] (12.15) A ˙ j k = 0 , ϕ = Φ a ¨ k x k +  constant.   Newtonian potential in   new coordinate system   Newtonian potential in   old coordinate system  {:[(12.15)A^(˙)_(j^(')k)=0","quadphi^(')=Phi-a^(¨)^(k)x^(k)+" constant. "],[[[" Newtonian potential in "],[" new coordinate system "]]]:}{:[[" Newtonian potential in "],[" old coordinate system "]]:}\begin{gather*} \dot{A}_{j^{\prime} k}=0, \quad \phi^{\prime}=\Phi-\ddot{a}^{k} x^{k}+\text { constant. } \tag{12.15}\\ {\left[\begin{array}{l} \text { Newtonian potential in } \\ \text { new coordinate system } \end{array}\right]} \end{gather*} \begin{aligned} & {\left[\begin{array}{l} \text { Newtonian potential in } \\ \text { old coordinate system } \end{array}\right]} \end{aligned}(12.15)A˙jk=0,ϕ=Φa¨kxk+ constant. [ Newtonian potential in  new coordinate system ][ Newtonian potential in  old coordinate system ]
These results can be restated in words: any two Galilean coordinate systems are related by (1) a time-independent rotation of the space grid (same rotation on each space slice), and (2) a time-dependent translation of the space grid (translation possibly different on different slices)
(12.16) x j = A j k x k + a j ( t ) [constant] [time-dependent] (12.16) x j = A j k x k + a j ( t )  [constant]   [time-dependent]  {:[(12.16)x^(j^('))=A_(j^(')k)x^(k)+a^(j^(')(t))],[" [constant] "uarr]:}[[],[" [time-dependent] "]\begin{gather*} x^{j^{\prime}}=A_{j^{\prime} k} x^{k}+a^{j^{\prime}(t)} \tag{12.16}\\ \text { [constant] } \uparrow \end{gather*} \begin{array}{|c} \\ \text { [time-dependent] } \end{array}(12.16)xj=Ajkxk+aj(t) [constant]  [time-dependent] 
The Newtonian potential is not a function defined in spacetime with existence independent of all coordinate systems. (There is no coordinate-free way to measure it.) Rather, it depends for its existence on a particular choice of Galilean coordinates; and if the choice is changed via equation (12.16), then Φ Φ Phi\PhiΦ is changed:
(12.17) Φ = Φ a ¨ k x k (12.17) Φ = Φ a ¨ k x k {:(12.17)Phi^(')=Phi-a^(¨)^(k)x^(k):}\begin{equation*} \Phi^{\prime}=\Phi-\ddot{a}^{k} x^{k} \tag{12.17} \end{equation*}(12.17)Φ=Φa¨kxk
(By contrast, an existence independent of all coordinates is granted to the universal time t ( P ) t ( P ) t(P)t(\mathscr{P})t(P) and the covariant derivative grad\boldsymbol{\nabla}.)
Were all the matter in the universe concentrated in a finite region of space and surrounded by emptiness ("island universe"), then one could impose the global boundary condition
(12.18) Φ 0 as r ( x k x k ) 1 / 2 (12.18) Φ 0  as  r x k x k 1 / 2 {:(12.18)Phi longrightarrow0" as "r-=(x^(k)x^(k))^(1//2)longrightarrow oo:}\begin{equation*} \Phi \longrightarrow 0 \text { as } r \equiv\left(x^{k} x^{k}\right)^{1 / 2} \longrightarrow \infty \tag{12.18} \end{equation*}(12.18)Φ0 as r(xkxk)1/2
This would single out a subclass of Galilean coordinates ("absolute" Galilean coordinates), with a unique, common Newtonian potential. The transformation from one absolute Galilean coordinate system to any other would be

("Galilean transformation"). But, (1) by no local measurements could one ever distinguish these absolute Galilean coordinate systems from the broader class of Galilean systems (to distinguish, one must integrate the locally measurable quantity Φ , j = Γ j 00 Φ , j = Γ j 00 Phi_(,j)=Gamma^(j)_(00)\Phi_{, j}=\Gamma^{j}{ }_{00}Φ,j=Γj00 out to infinity); and (2) astronomical data deny that the real universe is an island of matter surrounded by emptiness.
It is instructive to compare Galilean coordinates and Newtonian spacetime as described above with Lorentz coordinates and the Minkowskii spacetime of special relativity, and with the general coordinates and Einstein spacetime of general relativity; see Boxes 12.2 and 12.3.
Transformations linking Galilean coordinate systems
Newtonian potential depends on choice of Galilean coordinate system
Absolute Galilean coordinates defined
Transformations linking absolute Galilean coordinate systems
Box 12.2 NEWTONIAN SPACETIME, MINKOWSKIIAN SPACETIME, AND EINSTEINIAN SPACETIME:
Query Newtonian spacetime Minkowskiian spacetime (special relativity) Einsteinian spacetime (general relativity)
What a priori geometric structures does spacetime possess?
(1) Universal time function t t ttt
(2) Covariant derivative
(3) Spatial metric ":"; but spacetime metric can not be defined (exercise 12.10)
(1) Universal time function t (2) Covariant derivative (3) Spatial metric ":"; but spacetime metric can not be defined (exercise 12.10)| (1) Universal time function $t$ | | :--- | | (2) Covariant derivative | | (3) Spatial metric ":"; but spacetime metric can not be defined (exercise 12.10) |
A spacetime metric that is flat (vanishing Riemann curvature) A spacetime metric
What preferred coordinate systems are present?
(1) Galilean coordinates in general
(2) Absolute Galilean coordinates in an island universe (this case not considered here)
(1) Galilean coordinates in general (2) Absolute Galilean coordinates in an island universe (this case not considered here)| (1) Galilean coordinates in general | | :--- | | (2) Absolute Galilean coordinates in an island universe (this case not considered here) |
Lorentz coordinates In general, every coordinate system is equally preferred (though in special cases with symmetry there are special preferred coordinates)
What is required to select out a particular preferred coordinate system?
(1) A single spatial orientation, the same throughout all spacetime (three Euler angles)
(2) The arbitrary world line of the origin of space coordinates (three functions of time)
(1) A single spatial orientation, the same throughout all spacetime (three Euler angles) (2) The arbitrary world line of the origin of space coordinates (three functions of time)| (1) A single spatial orientation, the same throughout all spacetime (three Euler angles) | | :--- | | (2) The arbitrary world line of the origin of space coordinates (three functions of time) |
(1) A single spatial orientation, the same throughout all spacetime (three Euler angles)
(2) The location of the origin of coordinates (four numbers)
(3) The velocity of the origin of space coordinates (three numbers)
(1) A single spatial orientation, the same throughout all spacetime (three Euler angles) (2) The location of the origin of coordinates (four numbers) (3) The velocity of the origin of space coordinates (three numbers)| (1) A single spatial orientation, the same throughout all spacetime (three Euler angles) | | :--- | | (2) The location of the origin of coordinates (four numbers) | | (3) The velocity of the origin of space coordinates (three numbers) |
All four functions of position x a ( P ) x a ( P ) x^(a)(P)x^{a}(\mathscr{P})xa(P)
Under what conditions is " 9 and 2 are simultaneous" well-defined? In general; it is a coordinate-free geometric concept Only after a choice of Lorentz frame has been made; "simultaneity" depends on the frame's velocity Only after arbitrary choice of time coordinate has been made
Under what conditions is " 9 and 2 occur at same point in space" welldefined? Only after choice of Galilean coordinates has been made Only after choice of Lorentz coordinates has been made Only after arbitrary choice of space coordinates has been made
Under what conditions is " u u u\boldsymbol{u}u and v v v\boldsymbol{v}v, at different events, point in same direction" well-defined? Only if u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are both spatial vectors ( d l , u = d , v = 0 ) ( d l , u = d , v = 0 ) ((:dl,u:)=(:d,v:)=0)(\langle\boldsymbol{d} l, \boldsymbol{u}\rangle=\langle\boldsymbol{d}, \boldsymbol{v}\rangle=0)(dl,u=d,v=0); or if they lie in the same space slice and are arbitrary vectors; or if there exists a preferred route connecting their locations, along which to compare them by parallel transport Always Only if u u u\boldsymbol{u}u and v v v\boldsymbol{v}v lie at events infinitesimally close together; or if there exists a preferred route (e.g., a unique geodesic) connecting their locations, along which to compare them by parallel transport
Under what conditions is "the invariant distance between q q q\mathscr{\mathscr { q }}q and Q Q Q^('')\mathscr{Q}^{\prime \prime}Q well-defined? Only if P P P\mathscr{P}P and Q Q Q\mathcal{Q}Q lie in the same space slice Always Only if Q Q Q\mathscr{Q}Q and 2 are sufficiently close together; or if there exists a unique preferred world line (e.g.,
Query Newtonian spacetime Minkowskiian spacetime (special relativity) Einsteinian spacetime (general relativity) What a priori geometric structures does spacetime possess? "(1) Universal time function t (2) Covariant derivative (3) Spatial metric ":"; but spacetime metric can not be defined (exercise 12.10)" A spacetime metric that is flat (vanishing Riemann curvature) A spacetime metric What preferred coordinate systems are present? "(1) Galilean coordinates in general (2) Absolute Galilean coordinates in an island universe (this case not considered here)" Lorentz coordinates In general, every coordinate system is equally preferred (though in special cases with symmetry there are special preferred coordinates) What is required to select out a particular preferred coordinate system? "(1) A single spatial orientation, the same throughout all spacetime (three Euler angles) (2) The arbitrary world line of the origin of space coordinates (three functions of time)" "(1) A single spatial orientation, the same throughout all spacetime (three Euler angles) (2) The location of the origin of coordinates (four numbers) (3) The velocity of the origin of space coordinates (three numbers)" All four functions of position x^(a)(P) Under what conditions is " 9 and 2 are simultaneous" well-defined? In general; it is a coordinate-free geometric concept Only after a choice of Lorentz frame has been made; "simultaneity" depends on the frame's velocity Only after arbitrary choice of time coordinate has been made Under what conditions is " 9 and 2 occur at same point in space" welldefined? Only after choice of Galilean coordinates has been made Only after choice of Lorentz coordinates has been made Only after arbitrary choice of space coordinates has been made Under what conditions is " u and v, at different events, point in same direction" well-defined? Only if u and v are both spatial vectors ((:dl,u:)=(:d,v:)=0); or if they lie in the same space slice and are arbitrary vectors; or if there exists a preferred route connecting their locations, along which to compare them by parallel transport Always Only if u and v lie at events infinitesimally close together; or if there exists a preferred route (e.g., a unique geodesic) connecting their locations, along which to compare them by parallel transport Under what conditions is "the invariant distance between q and Q^('') well-defined? Only if P and Q lie in the same space slice Always Only if Q and 2 are sufficiently close together; or if there exists a unique preferred world line (e.g.,| Query | Newtonian spacetime | Minkowskiian spacetime (special relativity) | Einsteinian spacetime (general relativity) | | :---: | :---: | :---: | :---: | | What a priori geometric structures does spacetime possess? | (1) Universal time function $t$ <br> (2) Covariant derivative <br> (3) Spatial metric ":"; but spacetime metric can not be defined (exercise 12.10) | A spacetime metric that is flat (vanishing Riemann curvature) | A spacetime metric | | What preferred coordinate systems are present? | (1) Galilean coordinates in general <br> (2) Absolute Galilean coordinates in an island universe (this case not considered here) | Lorentz coordinates | In general, every coordinate system is equally preferred (though in special cases with symmetry there are special preferred coordinates) | | What is required to select out a particular preferred coordinate system? | (1) A single spatial orientation, the same throughout all spacetime (three Euler angles) <br> (2) The arbitrary world line of the origin of space coordinates (three functions of time) | (1) A single spatial orientation, the same throughout all spacetime (three Euler angles) <br> (2) The location of the origin of coordinates (four numbers) <br> (3) The velocity of the origin of space coordinates (three numbers) | All four functions of position $x^{a}(\mathscr{P})$ | | Under what conditions is " 9 and 2 are simultaneous" well-defined? | In general; it is a coordinate-free geometric concept | Only after a choice of Lorentz frame has been made; "simultaneity" depends on the frame's velocity | Only after arbitrary choice of time coordinate has been made | | Under what conditions is " 9 and 2 occur at same point in space" welldefined? | Only after choice of Galilean coordinates has been made | Only after choice of Lorentz coordinates has been made | Only after arbitrary choice of space coordinates has been made | | Under what conditions is " $\boldsymbol{u}$ and $\boldsymbol{v}$, at different events, point in same direction" well-defined? | Only if $\boldsymbol{u}$ and $\boldsymbol{v}$ are both spatial vectors $(\langle\boldsymbol{d} l, \boldsymbol{u}\rangle=\langle\boldsymbol{d}, \boldsymbol{v}\rangle=0)$; or if they lie in the same space slice and are arbitrary vectors; or if there exists a preferred route connecting their locations, along which to compare them by parallel transport | Always | Only if $\boldsymbol{u}$ and $\boldsymbol{v}$ lie at events infinitesimally close together; or if there exists a preferred route (e.g., a unique geodesic) connecting their locations, along which to compare them by parallel transport | | Under what conditions is "the invariant distance between $\mathscr{\mathscr { q }}$ and $\mathscr{Q}^{\prime \prime}$ well-defined? | Only if $\mathscr{P}$ and $\mathcal{Q}$ lie in the same space slice | Always | Only if $\mathscr{Q}$ and 2 are sufficiently close together; or if there exists a unique preferred world line (e.g., |

Box 12.3 NEWTONIAN GRAVITY Á LA CARTAN, AND EINSTEINIAN GRAVITY: COMPARISON AND CONTRAST

Property Newton-Cartan Einstein
Idea in brief (formulations of the equivalence principle of very different scope) Laws of motion of free particles in a local, freely falling, nonrotating frame are identical to Newton's laws of motion as expressed in a gravity-free Galilean frame Laws of physics in a local, freely falling, nonrotating frame are identical with the laws of physics as formulated in special relativity in a Lorentz frame
Idea even more briefly stated Point mechanics simple in a local inertial frame Everything simple in a local inertial frame
Consequence (tested to one part in 10 11 10 11 10^(11)10^{11}1011 by Roll-KrotkovDicke experiment) Test particles of diverse composition started with same initial position and same initial velocity follow the same world line ("definition of geodesic") Test particles of diverse composition started with same initial position and same initial velocity follow the same world line ("definition of geodesic")
Another consequence In every local region, there exists a local frame ("freely falling frame") in which all geodesics appear straight (all Γ α k p = 0 Γ α k p = 0 Gamma^(alpha)_(kp)=0\Gamma^{\alpha}{ }_{k p}=0Γαkp=0 ) In every local region there exists a local frame ("freely falling frame") in which all geodesics appear straight (all Γ α μ ν = 0 Γ α μ ν = 0 Gamma^(alpha)_(mu nu)=0\Gamma^{\alpha}{ }_{\mu \nu}=0Γαμν=0 )
Property Newton-Cartan Einstein Idea in brief (formulations of the equivalence principle of very different scope) Laws of motion of free particles in a local, freely falling, nonrotating frame are identical to Newton's laws of motion as expressed in a gravity-free Galilean frame Laws of physics in a local, freely falling, nonrotating frame are identical with the laws of physics as formulated in special relativity in a Lorentz frame Idea even more briefly stated Point mechanics simple in a local inertial frame Everything simple in a local inertial frame Consequence (tested to one part in 10^(11) by Roll-KrotkovDicke experiment) Test particles of diverse composition started with same initial position and same initial velocity follow the same world line ("definition of geodesic") Test particles of diverse composition started with same initial position and same initial velocity follow the same world line ("definition of geodesic") Another consequence In every local region, there exists a local frame ("freely falling frame") in which all geodesics appear straight (all Gamma^(alpha)_(kp)=0 ) In every local region there exists a local frame ("freely falling frame") in which all geodesics appear straight (all Gamma^(alpha)_(mu nu)=0 )| Property | Newton-Cartan | Einstein | | :---: | :---: | :---: | | Idea in brief (formulations of the equivalence principle of very different scope) | Laws of motion of free particles in a local, freely falling, nonrotating frame are identical to Newton's laws of motion as expressed in a gravity-free Galilean frame | Laws of physics in a local, freely falling, nonrotating frame are identical with the laws of physics as formulated in special relativity in a Lorentz frame | | Idea even more briefly stated | Point mechanics simple in a local inertial frame | Everything simple in a local inertial frame | | Consequence (tested to one part in $10^{11}$ by Roll-KrotkovDicke experiment) | Test particles of diverse composition started with same initial position and same initial velocity follow the same world line ("definition of geodesic") | Test particles of diverse composition started with same initial position and same initial velocity follow the same world line ("definition of geodesic") | | Another consequence | In every local region, there exists a local frame ("freely falling frame") in which all geodesics appear straight (all $\Gamma^{\alpha}{ }_{k p}=0$ ) | In every local region there exists a local frame ("freely falling frame") in which all geodesics appear straight (all $\Gamma^{\alpha}{ }_{\mu \nu}=0$ ) |
Consequence of way light rays travel in real physical world?
Summary of spacetime structure
Disregarded or evaded. All light rays have same velocity? Speed depend on motion of source? Speed depend on motion of observer? Possible to move fast enough to catch up with a light ray? No satisfactory position on any of these issues
This structure expressed in mathematical language
Stratified into spacelike slices; geometry in each slice Euclidean; each slice characterized by value of universal time (geodesic parameter); displacement of one slice with respect to another not specified; no such thing as a spacetime interval
Γ α μ ν Γ α μ ν Gamma^(alpha)_(mu nu)\Gamma^{\alpha}{ }_{\mu \nu}Γαμν 's, yes; spacetime metric g u v g u v g_(uv)g_{u v}guv, no;
Γ i 00 = Φ x i ( i = 1 , 2 , 3 ) ; all other Γ α μ ν vanish Γ i 00 = Φ x i ( i = 1 , 2 , 3 ) ;  all other  Γ α μ ν  vanish  {:[Gamma^(i)_(00)=(del Phi)/(delx^(i))(i=1","2","3);],[" all other "Gamma^(alpha)_(mu nu)" vanish "]:}\begin{aligned} & \Gamma^{i}{ }_{00}=\frac{\partial \Phi}{\partial x^{i}}(i=1,2,3) ; \\ & \text { all other } \Gamma^{\alpha}{ }_{\mu \nu} \text { vanish } \end{aligned}Γi00=Φxi(i=1,2,3); all other Γαμν vanish 
Spacetime always and everywhere has local Lorentz character
No stratification. Welldefined interval between every event and every nearby event; spacetime has everywhere local Lorentz character, with one local frame (specific space and time axes) as good as another (other space and time axes); "homogeneous" rather than stratified
Γ α u p Γ α u p Gamma^(alpha)_(up)\Gamma^{\alpha}{ }_{u p}Γαup,s have no independent existence; all derived from
Γ μ ν α = g α β 1 2 ( g β ν x μ + g β ν x ν g μ ν x β ) Γ μ ν α = g α β 1 2 g β ν x μ + g β ν x ν g μ ν x β {:[Gamma_(mu nu)^(alpha)=g^(alpha beta)(1)/(2)((delg_(beta nu))/(delx^(mu)):}],[{:+(delg_(beta nu))/(delx^(nu))-(delg_(mu nu))/(delx^(beta)))]:}\begin{aligned} \Gamma_{\mu \nu}^{\alpha}= & g^{\alpha \beta} \frac{1}{2}\left(\frac{\partial g_{\beta \nu}}{\partial x^{\mu}}\right. \\ & \left.+\frac{\partial g_{\beta \nu}}{\partial x^{\nu}}-\frac{\partial g_{\mu \nu}}{\partial x^{\beta}}\right) \end{aligned}Γμνα=gαβ12(gβνxμ+gβνxνgμνxβ)
("metric theory of gravity")

EXERCISES

Figure 12.1.
The coordinate system carried by an orbital laboratory as it moves in a circular orbit about the Earth.

Exercise 12.4. CONNECTION COEFFICIENTS FOR ROTATING, ACCELERATING COORDINATES

Beginning with equation (12.4) for the connection coefficients of a Galilean coordinate system { x α ( P ) } x α ( P ) {x^(alpha)(P)}\left\{x^{\alpha}(\mathscr{P})\right\}{xα(P)}, derive the connection coefficients (12.14) of the coordinate system { x α ( P ) } x α ( P ) {x^(alpha)(P)}\left\{x^{\alpha}(\mathscr{P})\right\}{xα(P)} of equations (12.13). From this, verify that (12.15) are necessary and sufficient for { x α ( P ) } x α ( P ) {x^(alpha^('))(P)}\left\{x^{\alpha^{\prime}}(\mathscr{P})\right\}{xα(P)} to be Galilean.

Exercise 12.5. EINSTEIN'S ELEVATOR

Use the formalism of this chapter to discuss "Einstein's elevator"-i.e., the equivalence of "gravity" to an acceleration of one's reference frame. Which aspects of "gravity" are equivalent to an acceleration, and which are not?

Exercise 12.6. GEODESIC DEVIATION ABOVE THE EARTH

A manned orbital laboratory is put into a circular orbit about the Earth [radius of orbit = r 0 = r 0 =r_(0)=r_{0}=r0, angular velocity = ω = ( M / r 0 3 ) 1 / 2 = ω = M / r 0 3 1 / 2 =omega=(M//r_(0)^(3))^(1//2)-=\omega=\left(M / r_{0}^{3}\right)^{1 / 2}-=ω=(M/r03)1/2 why?]. An astronaut jetisons a bag of garbage and watches it move along its geodesic path. He observes its motion relative to (non-Galilean) space coordinates { x j ( P ) } x j ( P ) {x^(j)(P)}\left\{x^{j}(\mathscr{P})\right\}{xj(P)} which-see Figure 12.1-(1) are Euclidean at each moment of universal time [ ( / x j ) ( / x k ) = δ j k ] / x j / x k = δ j k [(del//delx^(j))*(del//delx^(k^(')))=delta_(jk)]\left[\left(\partial / \partial x^{j}\right) \cdot\left(\partial / \partial x^{k^{\prime}}\right)=\delta_{j k}\right][(/xj)(/xk)=δjk], (2) have origin at the laboratory's center, (3) have / x / x del//delx^(')\partial / \partial x^{\prime}/x pointing away from the Earth, (4) have / x / x del//delx^(')\partial / \partial x^{\prime}/x and / y / y del//dely^(')\partial / \partial y^{\prime}/y in the plane of orbit. Use the equation of geodesic deviation to calculate the motion of the garbage bag in this coordinate system. Verify the answer by examining the Keplerian orbits of laboratory and garbage. Hints: (1) Calculate R α β γ 8 R α β γ 8 R^(alpha^('))_(beta^(')gamma^(')8^('))R^{\alpha^{\prime}}{ }_{\beta^{\prime} \gamma^{\prime} 8^{\prime}}Rαβγ8 in this coordinate system by a trivial transformation of tensorial components. (2) Use equation (12.14) to calculate Γ α β γ Γ α β γ Gamma^(alpha^('))_(beta^(')gamma^('))\Gamma^{\alpha^{\prime}}{ }_{\beta^{\prime} \gamma^{\prime}}Γαβγ at the center of the laboratory (i.e., on the fiducial geodesic).

§12.4. GEOMETRIC, COORDINATE-FREE FORMULATION OF NEWTONIAN GRAVITY

To restate Newton's theory of gravity in coordinate-independent, geometric language is the principal goal of this chapter. It has been achieved, thus far, with extensive assistance from a special class of coordinate systems, the Galilean coordinates. To
climb out of Galilean coordinates and into completely coordinate-free language is straightforward in principle. One merely passes from index notation to abstract notation.
Example: Restate in coordinate-free language the condition Γ 0 α β = 0 Γ 0 α β = 0 Gamma^(0)_(alpha beta)=0\Gamma^{0}{ }_{\alpha \beta}=0Γ0αβ=0 of Galilean coordinates.
Solution: Write Γ 0 α β = β d t , e α Γ 0 α β = β d t , e α Gamma^(0)_(alpha beta)=-(:grad_(beta)dt,e_(alpha):)\Gamma^{0}{ }_{\alpha \beta}=-\left\langle\boldsymbol{\nabla}_{\beta} \boldsymbol{d} t, \boldsymbol{e}_{\alpha}\right\rangleΓ0αβ=βdt,eα; the vanishing of this for all α α alpha\alphaα means β d t = 0 β d t = 0 grad_(beta)dt=0\boldsymbol{\nabla}_{\beta} \boldsymbol{d} t=0βdt=0 for all β β beta\betaβ, which in turn means u d t = 0 u d t = 0 grad_(u)dt=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{d} t=0udt=0 for all u u u\boldsymbol{u}u. In words: the gradient of universal time is covariantly constant.
By this process one can construct a set of coordinate-free statements about Newtonian spacetime (Box 12.4) that are completely equivalent to the standard, nongeometric version of Newton's gravitation theory. From standard Newtonian theory, one can deduce these geometric statements (exercise 12.7); from these geometric statements, regarded as axioms, one can deduce standard Newtonian theory (exercise 12.8).
Coordinate-free, geometric axioms for Newton's theory of gravity

Exercise 12.7. FROM NEWTON TO CARTAN

From the standard axioms of Newtonian theory (last part of Box 12.4) derive the geometric axioms (first part of Box 12.4). Suggested procedure: Verify each of the geometric axioms by a calculation in the Galilean coordinate system. Make free use of the calculations and results in $ 12.1 $ 12.1 $12.1\$ 12.1$12.1.

Exercise 12.8. FROM CARTAN TO NEWTON

From the geometric axioms of Newtonian theory (first part of Box 12.4) derive the standard axioms (last part of Box 12.4). Suggested procedure: (1) Pick three orthonormal, spatial basis vectors ( e j e j e_(j)\boldsymbol{e}_{j}ej with e j e k = δ j k e j e k = δ j k e_(j)*e_(k)=delta_(jk)\boldsymbol{e}_{j} \cdot \boldsymbol{e}_{k}=\delta_{j k}ejek=δjk ) at some event P 0 P 0 P_(0)\mathscr{P}_{0}P0. Parallel transport each of them by arbitrary routes to all other events in spacetime.
(2) Use the condition R ( u , n ) e j = 0 R ( u , n ) e j = 0 R(u,n)e_(j)=0\mathscr{R}(\boldsymbol{u}, \boldsymbol{n}) \boldsymbol{e}_{j}=0R(u,n)ej=0 for all u u u\boldsymbol{u}u and n n n\boldsymbol{n}n [axiom (3)] and an argument like that in § 11.5 § 11.5 §11.5\S 11.5§11.5 to conclude: (a) the resultant vector fields e j e j e_(j)\boldsymbol{e}_{j}ej are independent of the arbitrary transport routes, (b) e j = 0 e j = 0 grade_(j)=0\boldsymbol{\nabla} \boldsymbol{e}_{j}=0ej=0 for the resultant fields, and (c) [ e j , e k ] = 0 e j , e k = 0 [e_(j),e_(k)]=0\left[\boldsymbol{e}_{j}, \boldsymbol{e}_{k}\right]=0[ej,ek]=0.
(3) Pick an arbitrary "time line", which passes through each space slice (slice of constant t t ttt ) once and only once. Parametrize it by t t ttt and select its tangent vector as the basis vector e 0 e 0 e_(0)\boldsymbol{e}_{0}e0 at each event along it. Parallel transport each of these e 0 e 0 e_(0)\boldsymbol{e}_{0}e0 's throughout its respective space slice by arbitrary routes.
(4) From axiom (4) conclude that the resultant field is independent of the transport routes; also show that the above construction process guarantees j e 0 = 0 e j = 0 j e 0 = 0 e j = 0 grad_(j)e_(0)=grad_(0)e_(j)=0\boldsymbol{\nabla}_{j} \boldsymbol{e}_{0}=\boldsymbol{\nabla}_{0} \boldsymbol{e}_{j}=0je0=0ej=0.
(5) Show that [ e α , e β ] = 0 e α , e β = 0 [e_(alpha),e_(beta)]=0\left[\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right]=0[eα,eβ]=0 for all pairs of the four basis-vector fields, and conclude from this that there exists a coordinate system ("Galilean coordinates") in which e α = / x α e α = / x α e_(alpha)=del//delx^(alpha)\boldsymbol{e}_{\alpha}=\partial / \partial x^{\alpha}eα=/xα (see §11.5 and exercise 9.9).
(6) Show that in this coordinate system e j e k = δ j k e j e k = δ j k e_(j)*e_(k)=delta_(jk)\boldsymbol{e}_{j} \cdot \boldsymbol{e}_{k}=\delta_{j k}ejek=δjk everywhere (space coordinates are Euclidean), and the only nonzero components of the connection coefficient are Γ j 00 Γ j 00 Gamma^(j)_(00)\Gamma^{j}{ }_{00}Γj00; here axioms (6) and (2) will be helpful.
(7) From the self-adjoint property of the Jacobi curvature operator (axiom 7) show that R j 0 k 0 = R k 0 j 0 R j 0 k 0 = R k 0 j 0 R^(j)_(0k0)=R^(k)_(0j0)R^{j}{ }_{0 k 0}=R^{k}{ }_{0 j 0}Rj0k0=Rk0j0; show that in terms of the connection coefficients this reads Γ j 00 , k = Γ k 00 , j Γ j 00 , k = Γ k 00 , j Gamma^(j)_(00,k)=Gamma^(k)_(00,j)\Gamma^{j}{ }_{00, k}=\Gamma^{k}{ }_{00, j}Γj00,k=Γk00,j; and from this conclude that there exists a potential Φ Φ Phi\PhiΦ such that Γ j 00 = Φ , j Γ j 00 = Φ , j Gamma^(j)_(00)=Phi_(,j)\Gamma^{j}{ }_{00}=\Phi_{, j}Γj00=Φ,j.
(8) Show that the geometric field equation (axiom 5) reduces to Poisson's equation 2 Φ = 4 π ρ 2 Φ = 4 π ρ grad^(2)Phi=4pi rho\nabla^{2} \boldsymbol{\Phi}=4 \pi \rho2Φ=4πρ.
(9) Show that the geodesic equation for free fall (axiom 8) reduces to the Newtonian equation of motion d 2 x j / d t 2 + Φ j = 0 d 2 x j / d t 2 + Φ j = 0 d^(2)x^(j)//dt^(2)+Phi_(j)=0d^{2} x^{j} / d t^{2}+\Phi_{j}=0d2xj/dt2+Φj=0.

EXERCISES

Box 12.4 NEWTONIAN GRAVITY: GEOMETRIC FORMULATION CONTRASTED WITH STANDARD FORMULATION

Geometric Formulation

Newton's theory of gravity and the properties of Newtonian spacetime can be derived from the following axioms. (For derivation see exercise 12.8.)
(1) There exists a function t t ttt called "universal time", and a symmetric covariant derivative grad\mathbf{\nabla} (with associated geodesics, parallel transport law, curvature operator, etc.).
(2) The 1 -form d t d t dt\boldsymbol{d} tdt is covariantly constant; i.e.,
u d t = 0 for all u . u d t = 0  for all  u . grad_(u)dt=0" for all "u.\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{d} t=0 \text { for all } \boldsymbol{u} .udt=0 for all u.
[Consequence: if w w w\boldsymbol{w}w is a spatial vector field (i.e., w w w\boldsymbol{w}w lies everywhere in a surface of constant t t ttt; i.e. d t , w = 0 d t , w = 0 (:dt,w:)=0\langle\boldsymbol{d} t, \boldsymbol{w}\rangle=0dt,w=0 everywhere), then u w u w grad_(u)w\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{w}uw is also spatial for every u u u\boldsymbol{u}u,

(3) Spatial vectors are unchanged by parallel transport around infinitesimal closed curves; i.e.,
( u , n ) w = 0 if w is spatial, for every u and n . ( u , n ) w = 0  if  w  is spatial, for every  u  and  n . ℜ(u,n)w=0" if "w" is spatial, for every "u" and "n.\Re(\boldsymbol{u}, \boldsymbol{n}) \boldsymbol{w}=0 \text { if } \boldsymbol{w} \text { is spatial, for every } \boldsymbol{u} \text { and } \boldsymbol{n} .(u,n)w=0 if w is spatial, for every u and n.
(4) All vectors are unchanged by parallel transport around infinitesimal, spatial, closed curves; i.e.,
R ( v , w ) = 0 for every spatial v and w . R ( v , w ) = 0  for every spatial  v  and  w . R(v,w)=0" for every spatial "v" and "w.\mathscr{R}(\boldsymbol{v}, \boldsymbol{w})=0 \text { for every spatial } \boldsymbol{v} \text { and } \boldsymbol{w} .R(v,w)=0 for every spatial v and w.
(5) The Ricci curvature tensor, R α β R μ α μ β R α β R μ α μ β R_(alpha beta)-=R^(mu)_(alpha mu beta)R_{\alpha \beta} \equiv R^{\mu}{ }_{\alpha \mu \beta}RαβRμαμβ, has the form
R i c c i = 4 π ρ d t d t R i c c i = 4 π ρ d t d t Ricci=4pi rho dt ox dt\boldsymbol{R i c c i}=4 \pi \rho \boldsymbol{d} t \otimes \boldsymbol{d} tRicci=4πρdtdt
where ρ ρ rho\rhoρ is the density of mass.
(6) There exists a metric "." defined on spatial vectors only, which is compatible with the covariant derivative in this sense: for any spatial w w w\boldsymbol{w}w and v v v\boldsymbol{v}v, and for any u u u\boldsymbol{u}u whatsoever,
u ( w v ) = ( u w ) v + w ( u v ) u ( w v ) = u w v + w u v grad_(u)(w*v)=(grad_(u)w)*v+w*(grad_(u)v)\nabla_{u}(\boldsymbol{w} \cdot v)=\left(\nabla_{u} w\right) \cdot v+w \cdot\left(\nabla_{u} v\right)u(wv)=(uw)v+w(uv)
[Note: axioms (1), (2), and (3) guarantee that such a spatial metric can exist; see exercise 12.9.]
(7) The Jacobi curvature operator G ( u , n ) G ( u , n ) G(u,n)\mathcal{G}(\boldsymbol{u}, \boldsymbol{n})G(u,n), defined for any vectors u , n , p u , n , p u,n,p\boldsymbol{u}, \boldsymbol{n}, \boldsymbol{p}u,n,p by
H ( u , n ) p = 1 2 [ R ( p , n ) u + R ( p , u ) n ] , H ( u , n ) p = 1 2 [ R ( p , n ) u + R ( p , u ) n ] , H(u,n)p=(1)/(2)[R(p,n)u+R(p,u)n],\mathscr{H}(\boldsymbol{u}, \boldsymbol{n}) \boldsymbol{p}=\frac{1}{2}[\mathscr{R}(\boldsymbol{p}, \boldsymbol{n}) \boldsymbol{u}+\mathscr{R}(\boldsymbol{p}, \boldsymbol{u}) \boldsymbol{n}],H(u,n)p=12[R(p,n)u+R(p,u)n],
is "self-adjoint" when operating on spatial vectors; i.e.,
v [ f ( u , n ) w ] = w [ f ( u , n ) v ] for all spatial v , w ; and for any u , n . v [ f ( u , n ) w ] = w [ f ( u , n ) v ]  for all spatial  v , w ;  and for any  u , n . {:[v*[f(u","n)w]=w*[f(u","n)v]" for all spatial "v","w;],[" and for any "u","n.]:}\begin{aligned} & \boldsymbol{v} \cdot[\mathscr{f}(\boldsymbol{u}, \boldsymbol{n}) \boldsymbol{w}]=\boldsymbol{w} \cdot[\mathcal{f}(\boldsymbol{u}, \boldsymbol{n}) \boldsymbol{v}] \text { for all spatial } \boldsymbol{v}, \boldsymbol{w} ; \\ & \text { and for any } \boldsymbol{u}, \boldsymbol{n} . \end{aligned}v[f(u,n)w]=w[f(u,n)v] for all spatial v,w; and for any u,n.
(8) "Ideal rods" measure the lengths that are calculated with the spatial metric; "ideal clocks" measure universal time t t ttt (or some multiple thereof); and "freely falling particles" move along geodesics of grad\boldsymbol{\nabla}. [Note: this can be regarded as a definition of "ideal rods," "ideal clocks," and "freely falling particles." A more complete theory (e.g., general relativity; see § 16.4 § 16.4 §16.4\S 16.4§16.4 ) would predict in advance whether a given physical rod or clock is ideal, and whether a given real particle is freely falling.]
Note: For an alternative but equivalent set of axioms, see pp. 106-107 of Trautman (1965).

Standard Formulation

The following standard axioms are equivalent to the above.
(1) There exist a universal time t t ttt, a set of Cartesian space coordinates x j x j x^(j)x^{j}xj (called "Galilean coordinates"), and a Newtonian gravitational potential Φ Φ Phi\PhiΦ.
(2) The density of mass ρ ρ rho\rhoρ generates the Newtonian potential by Poisson's equation,
2 Φ 2 Φ x j x j = 4 π ρ . 2 Φ 2 Φ x j x j = 4 π ρ . grad^(2)Phi-=(del^(2)Phi)/(delx^(j)delx^(j))=4pi rho.\nabla^{2} \Phi \equiv \frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{j}}=4 \pi \rho .2Φ2Φxjxj=4πρ.
(3) The equation of motion for a freely falling particle is
d 2 x j d t 2 + Φ x j = 0 . d 2 x j d t 2 + Φ x j = 0 . (d^(2)x^(j))/(dt^(2))+(del Phi)/(delx^(j))=0.\frac{d^{2} x^{j}}{d t^{2}}+\frac{\partial \boldsymbol{\Phi}}{\partial x^{j}}=0 .d2xjdt2+Φxj=0.
(4) "Ideal rods" measure the Galilean coordinate lengths; "ideal clocks" measure universal time.

Exercise 12.9. SPATIAL METRIC ALLOWED BY OTHER AXIOMS

Show that the geometric axioms (1), (2), and (3) of Box 12.4 permit one to introduce a spatial metric satisfying axiom (6). Hint: Pick an arbitrary spatial basis { e i } e i {e_(i)}\left\{\boldsymbol{e}_{i}\right\}{ei} at some event. Define it to be orthonormal, e j e k δ j k e j e k δ j k e_(j)*e_(k)-=delta_(jk)\boldsymbol{e}_{j} \cdot \boldsymbol{e}_{k} \equiv \delta_{j k}ejekδjk. Extend this basis through all spacetime by the method used in (1) of exercise 12.8. Define e j e k δ j k e j e k δ j k e_(j)*e_(k)-=delta_(jk)\boldsymbol{e}_{j} \cdot \boldsymbol{e}_{k} \equiv \delta_{j k}ejekδjk everywhere in spacetime for this basis. Then prove that the resulting metric satisfies the compatibility condition of axiom (6).

Exercise 12.10. SPACETIME METRIC FORBIDDEN BY OTHER AXIOMS

Show that in Newtonian spacetime it is impossible to construct a nondegenerate spacetime metric g g g\boldsymbol{g}g, defined on all vectors, that is compatible with the covariant derivative in the sense that
(12.20) u g ( n , p ) = g ( u n , p ) + g ( n , u p ) (12.20) u g ( n , p ) = g u n , p + g n , u p {:(12.20)grad_(u)g(n","p)=g(grad_(u)n,p)+g(n,grad_(u)p):}\begin{equation*} \nabla_{u} g(n, p)=g\left(\nabla_{u} n, p\right)+g\left(n, \nabla_{u} p\right) \tag{12.20} \end{equation*}(12.20)ug(n,p)=g(un,p)+g(n,up)
Note: to prove this requires mastery of the material in Chapter 8 or 13; so study either 8 or 13 before tackling it. Hint: Assume that such a g g g\boldsymbol{g}g exists. Show, by the methods of exercise 12.8 , that in a Galilean coordinate system the spatial components g j k g j k g_(jk)g_{j k}gjk are independent of position in spacetime. Then use this and the form of R α β γ δ R α β γ δ R^(alpha)_(beta gamma delta)R^{\alpha}{ }_{\beta \gamma \delta}Rαβγδ in Galilean coordinates to prove R i 0 k 0 R i 0 k 0 R_(i0k0)R_{i 0 k 0}Ri0k0 and R 0 i k 0 R 0 i k 0 -R_(0ik0)-R_{0 i k 0}R0ik0 are not identical, a result that conflicts with the symmetries of the Riemann tensor [eq. (8.45)] in a manifold with compatible metric and covariant derivative.
The principle of general covariance has no forcible content
Twentieth-century viewpoint judges a theory by simplicity of its geometric formulation
Einstein's theory of gravity is simple; Newton's is complex

§12.5. THE GEOMETRIC VIEW OF PHYSICS: A CRITIQUE

An important digression is in order.
"Every physical quantity must be describable by a (coordinate-free) geometric object, and the laws of physics must all be expressible as geometric relationships between these geometric objects." This view of physics, sometimes known as the "principle of general covariance," pervades twentieth-century thinking. But does it have any forcible content? No, not at all, according to one viewpoint that dates back to Kretschmann (1917). Any physical theory originally written in a special coordinate system can be recast in geometric, coordinate-free language. Newtonian theory is a good example, with its equivalent geometric and standard formulations (Box 12.4). Hence, as a sieve for separating viable theories from nonviable theories, the principle of general covariance is useless.
But another viewpoint is cogent. It constructs a powerful sieve in the form of a slightly altered and slightly more nebulous principle: "Nature likes theories that are simple when stated in coordinate-free, geometric language."* According to this principle, Nature must love general relativity, and it must hate Newtonian theory. Of all theories ever conceived by physicists, general relativity has the simplest, most elegant geometric foundation [three axioms: (1) there is a metric; (2) the metric is governed by the Einstein field equation G = 8 π T G = 8 π T G=8pi T\boldsymbol{G}=8 \pi \boldsymbol{T}G=8πT; (3) all special relativistic laws of physics are valid in local Lorentz frames of metric]. By contrast, what diabolically
clever physicist would ever foist on man a theory with such-a complicated geometric foundation as Newtonian theory?
Of course, from the nineteenth-century viewpoint, the roles are reversed. It judged simplicity of theories by examining their coordinate formulations. In Galilean coordinates, Newtonian theory is beautifully simple. Expressed as differential equations for the metric coefficients in a specific coordinate system, Einstein's field equations (ten of them now!) are horrendously complex.
The geometric, twentieth-century view prevails because it accords best with experimental data (see Chapters 38 40 38 40 38-4038-403840 ). In Chapter 17 it will be applied ruthlessly to make Einstein's field equation seem compelling.
снаттен 13
RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF ALL

Abstract

Philosophy is written in this great book (by which I mean the universe) which stands always open to our view, but it cannot be understood unless one first learns how to comprehend the language and interpret the symbols in which it is written, and its symbols are triangles, circles, and other geometric figures, without which it is not humanly possible to comprehend even one word of it; without these one wanders in a dark labyrinth.

GALILEO GALILEI (1623)

§13.1. NEW FEATURES IMPOSED ON GEOMETRY BY LOCAL VALIDITY OF SPECIAL RELATIVITY

This chapter is entirely Track 2.
Chapters 9-11 are necessary preparation for it. It will be needed as preparation for
(1) the second half of Chapter 14 (calculation of curvature), and
(2) the details, but not the message, of Chapter 15 (Bianchi identities).
§ 13.6 (proper reference frame) will be useful throughout the applications of gravitation theory (Chapters 18-40).
Constraints imposed on spacetime by special relativity
Freely falling particles (geodesics) define and probe the structure of spacetime. This spacetime is curved. Gravitation is a manifestation of its curvature. So far, so good, in terms of Newton's theory of gravity as translated into geometric language by Cartan. What is absolutely unacceptable, however, is the further consequence of the Cartan-Newton viewpoint (Chapter 12): stratification of spacetime into slidable slices, with no meaning for the spacetime separation between an event in one slice and an event in another.
Of all the foundations of physics, none is more firmly established than special relativity; and of all the lessons of special relativity none stand out with greater force than these. (1) Spacetime, far from being stratified, is homogeneous and isotropic throughout any region small enough ("local region") that gravitational tide-producing effects ("spacetime curvatures") are negligible. (2) No local experiment whatsoever can distinguish one local inertial frame from another. (3) The speed of light is the same in every local inertial frame. (4) It is not possible to give frame-independent meaning to the separation in time ("no Newtonian stratifica-
tion"). (5) Between every event and every nearby event there exists a frameindependent, coordinate-independent spacetime interval ("Riemannian geometry"). (6) Spacetime is always and everywhere locally Lorentz in character ("local Lorentz character of this Riemannian geometry").
What mathematics gives all these physical properties? A metric; a metric that is locally Lorentz ( $ 813.2 $ 813.2 $813.2\$ 813.2$813.2 and 13.6). All else then follows. In particular, the metric destroys the stratified structure of Newtonian spacetime, as well as its gravitational potential and universal time coordinate. But not destroyed are the deepest features of Newtonian gravity: (1) the equivalence principle (as embodied in geodesic description of free-fall motion, § § 13.3 § § 13.3 §§13.3\S \S 13.3§§13.3 and 13.4); and (2) spacetime curvature (as measured by tidal effects, §13.5).
The skyscraper of vectors, forms, tensors (Chapter 9), geodesics, parallel transport, covariant derivative (Chapter 10), and curvature (Chapter 11) has rested on crumbling foundations-Newtonian physics and a geodesic law based on Newtonian physics. But with metric now on the scene, the whole skyscraper can be transferred to new foundations without a crack appearing. Only one change is necessary: the geodesic law must be selected in a new, relativistic way; a way based on metric ( § $ 13.3 § $ 13.3 §$13.3\S \$ 13.3§$13.3 and 13.4). Resting on metric foundations, spacetime curvature acquires additional and stronger properties (the skyscraper is redecorated and extended), which are studied in § 13.5 § 13.5 §13.5\S 13.5§13.5 and in Chapters 14 and 15 , and which lead almost inexorably to Einstein's field equation.

§13.2. METRIC

A spacetime metric; a curved spacetime metric; a locally Lorentz, curved spacetime metric. This is the foundation of spacetime geometry in the real, physical world. Therefore take a moment to recall what "metric" is in three contrasting languages.
In the language of elementary geometry, "metric" is a table giving the interval between every event and every other event (Box 13.1 and Figure 13.1). In the language of coordinates, "metric" is a set of ten functions of position, g μ ν ( x α ) g μ ν x α g_(mu nu)(x^(alpha))g_{\mu \nu}\left(x^{\alpha}\right)gμν(xα), such that the expression
(13.1) Δ s 2 = Δ τ 2 = g μ ν ( x α ) Δ x μ Δ x ν (13.1) Δ s 2 = Δ τ 2 = g μ ν x α Δ x μ Δ x ν {:(13.1)Deltas^(2)=-Deltatau^(2)=g_(mu nu)(x^(alpha))Deltax^(mu)Deltax^(nu):}\begin{equation*} \Delta s^{2}=-\Delta \tau^{2}=g_{\mu \nu}\left(x^{\alpha}\right) \Delta x^{\mu} \Delta x^{\nu} \tag{13.1} \end{equation*}(13.1)Δs2=Δτ2=gμν(xα)ΔxμΔxν
gives the interval between any event x α x α x^(alpha)x^{\alpha}xα and any nearby event x α + Δ x α x α + Δ x α x^(alpha)+Deltax^(alpha)x^{\alpha}+\Delta x^{\alpha}xα+Δxα. In the language of abstract differential geometry, metric is a bilinear machine, g ( ) g ( ) g-=(dots*dots)\boldsymbol{g} \equiv(\ldots \cdot \ldots)g(), to produce a number ["scalar product g ( u , v ) ( u v ) g ( u , v ) ( u v ) g(u,v)-=(u*v)\boldsymbol{g}(\boldsymbol{u}, \boldsymbol{v}) \equiv(\boldsymbol{u} \cdot \boldsymbol{v})g(u,v)(uv) "] out of two tangent vectors, u u u\boldsymbol{u}u and v v v\boldsymbol{v}v.
The link between the abstract, machine viewpoint and the concrete coordinate viewpoint is readily exhibited. Let the tangent vector
ξ Δ x α e α = Δ x α ( / x α ) ξ Δ x α e α = Δ x α / x α xi-=Deltax^(alpha)e_(alpha)=Deltax^(alpha)(del//delx^(alpha))\xi \equiv \Delta x^{\alpha} \boldsymbol{e}_{\alpha}=\Delta x^{\alpha}\left(\partial / \partial x^{\alpha}\right)ξΔxαeα=Δxα(/xα)
represent the displacement between two neighboring events. The abstract viewpoint gives
Δ s 2 ξ ξ g ( Δ x μ e μ , Δ x ν e ν ) = Δ x μ Δ x ν g ( e μ , e ν ) Δ s 2 ξ ξ g Δ x μ e μ , Δ x ν e ν = Δ x μ Δ x ν g e μ , e ν Deltas^(2)-=xi*xi-=g(Deltax^(mu)e_(mu),Deltax^(nu)e_(nu))=Deltax^(mu)Deltax^(nu)g(e_(mu),e_(nu))\Delta s^{2} \equiv \boldsymbol{\xi} \cdot \xi \equiv \boldsymbol{g}\left(\Delta x^{\mu} \boldsymbol{e}_{\mu}, \Delta x^{\nu} \boldsymbol{e}_{\nu}\right)=\Delta x^{\mu} \Delta x^{\nu} \boldsymbol{g}\left(\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right)Δs2ξξg(Δxμeμ,Δxνeν)=ΔxμΔxνg(eμ,eν)
Metric described in three languages

Box 13.1 METRIC DISTILLED FROM DISTANCES

Raw Data on Distances

Let the shape of the earth be described as in Figure 13.1, by giving distances between some of the principal identifiable points: buoys, ships, icebergs, lighthouses, peaks, and flags: points to a total of n = 2 × 10 7 n = 2 × 10 7 n=2xx10^(7)n=2 \times 10^{7}n=2×107. The total number of distances to be given is n ( n 1 ) / 2 = 2 × 10 14 n ( n 1 ) / 2 = 2 × 10 14 n(n-1)//2=2xx10^(14)n(n-1) / 2=2 \times 10^{14}n(n1)/2=2×1014. With 200 distances per page of printout, this means

10 12 10 12 10^(12)10^{12}1012 pages weighing 6 g each, or 6 × 10 6 6 × 10 6 6xx10^(6)6 \times 10^{6}6×106 metric tons of data. With 6 tons per truck this means 10 6 10 6 10^(6)10^{6}106 truckloads of data; or with one truck passing by every 5 seconds, 5 × 10 6 5 × 10 6 5xx10^(6)5 \times 10^{6}5×106 seconds or 2 months of night and day traffic to get in the data.

First Distillation: Distances to Nearby Points Only

Get distances between faraway points by adding distances covered on the elementary short legs of the trip. Boil down the table of distances to give only the distance between each point and the hundred nearest points. Now have 100 n = 2 × 10 9 100 n = 2 × 10 9 100 n=2xx10^(9)100 n=2 \times 10^{9}100n=2×109 distances, or 2 × 10 9 / 200 = 10 7 2 × 10 9 / 200 = 10 7 2xx10^(9)//200=10^(7)2 \times 10^{9} / 200=10^{7}2×109/200=107 pages of data, or 60 tons of records, or 10 truckloads.

Second Distillation: Distances Between Nearby Points in Terms of Metric

Idealize the surface of the earth as smooth. Then in any sufficiently limited region the geometry is Euclidean. This circumstance has a happy consequence. It is enough to know a few distances between the nearby points to be able to determine all the

distances between the nearby points. Locate point 2 so that (102) is a right triangle; thus, ( 12 ) 2 = ( 10 ) 2 + ( 20 ) 2 ( 12 ) 2 = ( 10 ) 2 + ( 20 ) 2 (12)^(2)=(10)^(2)+(20)^(2)(12)^{2}=(10)^{2}+(20)^{2}(12)2=(10)2+(20)2. Consider a point 3 close to 0 . Define
x ( 3 ) = ( 13 ) ( 10 ) x ( 3 ) = ( 13 ) ( 10 ) x(3)=(13)-(10)x(3)=(13)-(10)x(3)=(13)(10)
and
y ( 3 ) = ( 23 ) ( 20 ) . y ( 3 ) = ( 23 ) ( 20 ) . y(3)=(23)-(20).y(3)=(23)-(20) .y(3)=(23)(20).
Then the distance (03) does not have to be supplied independently; it can be calculated from the formula*
( 03 ) 2 = [ x ( 3 ) ] 2 + [ y ( 3 ) ] 2 . ( 03 ) 2 = [ x ( 3 ) ] 2 + [ y ( 3 ) ] 2 . (03)^(2)=[x(3)]^(2)+[y(3)]^(2).(03)^{2}=[x(3)]^{2}+[y(3)]^{2} .(03)2=[x(3)]2+[y(3)]2.
Similarly for a point 4 and its distance (04) from the local origin 0 . Similarly for the distance ( m n ) ( m n ) (mn)(m n)(mn) between any two points m m mmm and n n nnn that are close to 0 :
( m n ) 2 = [ x ( m ) x ( n ) ] 2 + [ y ( m ) y ( n ) ] 2 . ( m n ) 2 = [ x ( m ) x ( n ) ] 2 + [ y ( m ) y ( n ) ] 2 . (mn)^(2)=[x(m)-x(n)]^(2)+[y(m)-y(n)]^(2).(m n)^{2}=[x(m)-x(n)]^{2}+[y(m)-y(n)]^{2} .(mn)2=[x(m)x(n)]2+[y(m)y(n)]2.
Thus it is only needful to have the distance ( 1 m ) (from point 1 ) and ( 2 m 2 m 2m2 m2m ) (from point 2 ) for each point m m mmm close to 0 ( m = 3 , 4 , , N + 2 ) 0 ( m = 3 , 4 , , N + 2 ) 0(m=3,4,dots,N+2)0(m=3,4, \ldots, N+2)0(m=3,4,,N+2) to be able to work out
Box 13.1 (continued)
its distance from every point n n nnn close to 0 . The prescription to determine the N ( N 1 ) / 2 N ( N 1 ) / 2 N(N-1)//2N(N-1) / 2N(N1)/2 distances between these N N NNN nearby points can be reexpressed to advantage in these words: (1) each point has two coordinates, x x xxx and y y yyy; and (2) the distance is given in terms of these coordinates by the standard Euclidean metric; thus
( Δ s ) 2 = ( Δ x ) 2 + ( Δ y ) 2 ( Δ s ) 2 = ( Δ x ) 2 + ( Δ y ) 2 (Delta s)^(2)=(Delta x)^(2)+(Delta y)^(2)(\Delta s)^{2}=(\Delta x)^{2}+(\Delta y)^{2}(Δs)2=(Δx)2+(Δy)2
Having gone this far on the basis of "distance geometry" (for more on which, see Robb 1914 and 1936), one can generalize from a small region (Euclidean) to a large region (not Euclidean). Introduce any arbitrary smooth pair of every-where-independent curvilinear coordinates x k x k x^(k)x^{k}xk, and express distance, not only in the immediate neighborhood of the point 0 , but also in the immediate neighborhood of every point of the surface (except places where one has to go to another coordinate patch; at least two patches needed for 2 -sphere) in terms of the formula
d s 2 = g i k d x j d x k d s 2 = g i k d x j d x k ds^(2)=g_(ik)dx^(j)dx^(k)d s^{2}=g_{i k} d x^{j} d x^{k}ds2=gikdxjdxk
Thus out of the table of distances between nearby points one has distilled now five numbers per point (two coordinates, x 1 , x 2 x 1 , x 2 x^(1),x^(2)x^{1}, x^{2}x1,x2, and three metric coefficients, g 11 , g 12 = g 11 , g 12 = g_(11),g_(12)=g_{11}, g_{12}=g11,g12= g 21 g 21 g_(21)g_{21}g21, and g 22 g 22 g_(22)g_{22}g22 ), down by a factor of 100 / 5 = 20 100 / 5 = 20 100//5=20100 / 5=20100/5=20 from what one had before (now 3 tons of data, or half a truckload).

Third Distillation: Metric Coefficients Expressed as Analytical Functions of Coordinates

Instead of giving the three metric coefficients at each of the 2 × 10 7 2 × 10 7 2xx10^(7)2 \times 10^{7}2×107 points of the surface, give them as functions of the two coordinates x 1 , x 2 x 1 , x 2 x^(1),x^(2)x^{1}, x^{2}x1,x2, in terms of a power series or an expansion in spherical harmonics or otherwise with some modest number, say 100 , of adjustable coefficients. Then the information about the geometry itself (as distinct from the coordinates of the 2 × 10 7 2 × 10 7 2xx10^(7)2 \times 10^{7}2×107 points located on that geometry) is caught up in these three hundred coefficients, a single page of printout. Goodbye to any truck! In brief, metric provides a shorthand way of giving the distance between every point and every other point-but its role, its justification and its meaning lies in these distances and only in these many distances.

Capetown
Figure 13.1.
Distances determine geometry. Upper left: Sufficiently great tidal forces, applied to the earth with tailored timing, have deformed it to the shape of a tear drop. Lower left: This tear drop is approximated by a polyhedron built out of triangles ("skeleton geometry"). The approximation can be made arbitrarily good by making the number of triangles sufficiently great and the size of each sufficiently small. Upper right: The geometry in each triangle is Euclidean: giving the three edge lengths fixes all the features of the figure, including the indicated angle. Lower right: The triangles that belong to a given vertex, laid out on a flat surface, fail to meet. The deficit angle measures the amount of curvature concentrated at that vertex on the tear-drop earth. The sum of these deficit angles for all vertices of the tear drop equals 4 π 4 π 4pi4 \pi4π. This "Gauss-Bonnet theorem" is valid for any figure with the topology of the 2 -sphere; for the simplest figure of all, a tetrahedron, four vertices with a deficit angle at each of 180 180 180^(@)180^{\circ}180 are needed- 3 triangles × 60 × 60 xx60^(@)\times 60^{\circ}×60 per triangle available = 180 = 180 =180^(@)=180^{\circ}=180 deficit. In brief, the shape of the tear drop, in the given skeleton-geometry approximation, is determined by its 50 visible edge lengths plus, say, 32 more edge lengths hidden behind the figure, or a total of 82 edge lengths, and by nothing more ("distances determine geometry"). "Metric" tells the distance between every point and every nearby point. If volcanic action raises Rejkjavik, the distances between that Icelandic capital and nearby points increase accordingly; distances again reveal shape. Conversely, that there is not a great bump on the earth in the vicinity of Iceland, and that the earth does not now have a tear-drop shape, can be unambiguously established by analyzing the pattern of distances from point to point in a sufficiently well-distributed network of points, with no call for any observations other than measurements of distance.
Covariant components of metric
"Line element" compared with "metric as bilinear machine"
Metric produces a correspondence between 1 -forms and tangent vectors
for the interval between those events; comparison with the coordinate viewpoint [equation (13.1)] reveals
(13.2) g μ ν = g ( e μ , e ν ) = e μ e ν (13.2) g μ ν = g e μ , e ν = e μ e ν {:(13.2)g_(mu nu)=g(e_(mu),e_(nu))=e_(mu)*e_(nu):}\begin{equation*} g_{\mu \nu}=\boldsymbol{g}\left(\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right)=\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu} \tag{13.2} \end{equation*}(13.2)gμν=g(eμ,eν)=eμeν
(standard equation for calculating components of a tensor).
Just as modern differential geometry replaces the old style "differential" d f d f dfd fdf by the "differential form" d f d f df\boldsymbol{d} fdf (Box 2.3 , page 63 ), so it also replaces the old-style "line element"
(13.3) d s 2 = g μ ν d x μ d x ν = ( "interval between x α and x α + d x α " ) (13.3) d s 2 = g μ ν d x μ d x ν =  "interval between  x α  and  x α + d x α " {:(13.3)ds^(2)=g_(mu nu)dx^(mu)dx^(nu)=(" "interval between "x^(alpha)" and "x^(alpha)+dx^(alpha")):}\begin{equation*} d s^{2}=g_{\mu \nu} d x^{\mu} d x^{\nu}=\left(\text { "interval between } x^{\alpha} \text { and } x^{\alpha}+d x^{\alpha "}\right) \tag{13.3} \end{equation*}(13.3)ds2=gμνdxμdxν=( "interval between xα and xα+dxα")
by the bilinear machine ("metric tensor")
(13.4) g d s 2 g μ ν d x μ d x ν (13.4) g d s 2 g μ ν d x μ d x ν {:(13.4)g-=ds^(2)-=g_(mu nu)dx^(mu)ox dx^(nu):}\begin{equation*} \boldsymbol{g} \equiv \boldsymbol{d} \boldsymbol{s}^{2} \equiv g_{\mu \nu} \boldsymbol{d} x^{\mu} \otimes \boldsymbol{d} x^{\nu} \tag{13.4} \end{equation*}(13.4)gds2gμνdxμdxν
The output g ( ξ , ξ ) g ( ξ , ξ ) g(xi,xi)\boldsymbol{g}(\boldsymbol{\xi}, \boldsymbol{\xi})g(ξ,ξ) of this machine, for given displacement-vector input, is identical to the old-style interval. Hence, d s 2 = g μ ν d x μ d x ν d s 2 = g μ ν d x μ d x ν ds^(2)=g_(mu nu)dx^(mu)ox dx^(nu)\boldsymbol{d} \boldsymbol{s}^{2}=g_{\mu \nu} \boldsymbol{d} x^{\mu} \otimes \boldsymbol{d} x^{\nu}ds2=gμνdxμdxν represents the interval of an unspecified displacement; and the act of inserting ξ ξ xi\boldsymbol{\xi}ξ into the slots of d s 2 d s 2 ds^(2)\boldsymbol{d} \boldsymbol{s}^{2}ds2 is the act of making explicit the interval g ( ξ , ξ ) = g μ ν Δ x μ Δ x ν g ( ξ , ξ ) = g μ ν Δ x μ Δ x ν g(xi,xi)=g_(mu nu)Deltax^(mu)Deltax^(nu)\boldsymbol{g}(\boldsymbol{\xi}, \boldsymbol{\xi})=g_{\mu \nu} \Delta x^{\mu} \Delta x^{\nu}g(ξ,ξ)=gμνΔxμΔxν of an explicit displacement.
In curved spacetime with metric, just as in flat spacetime with metric (§2.5), a particular 1 -form u ~ u ~ widetilde(u)\widetilde{\boldsymbol{u}}u~ corresponds to any given tangent vector u u u\boldsymbol{u}u :
(13.5) u ~ is defined by " u ~ , v g ( u , v ) for all v " (13.5) u ~  is defined by "  u ~ , v g ( u , v )  for all  v  "  {:(13.5) widetilde(u)" is defined by " " widetilde(u)","v:)-=g(u","v)" for all "v" " ":}\begin{equation*} \widetilde{\boldsymbol{u}} \text { is defined by " } \widetilde{\boldsymbol{u}}, \boldsymbol{v}\rangle \equiv \boldsymbol{g}(\boldsymbol{u}, \boldsymbol{v}) \text { for all } \boldsymbol{v} \text { " } \tag{13.5} \end{equation*}(13.5)u~ is defined by " u~,vg(u,v) for all v " 
("representation of the same physical quantity in the two alternative versions of vector and 1-form"; "corresponding representations" as ( 1 0 ) ( 1 0 ) ((1)/(0))\binom{1}{0}(10)-tensor and as ( 0 1 ) ( 0 1 ) ((0)/(1))\binom{0}{1}(01)-tensor). Example: the 1-form u ~ u ~ widetilde(u)\widetilde{\boldsymbol{u}}u~ corresponding to a basis vector u = e α u = e α u=e_(alpha)\boldsymbol{u}=\boldsymbol{e}_{\alpha}u=eα has components

thus
(13.6) g α β ω β is the 1 -form e ~ α corresponding to e α (13.6) g α β ω β  is the  1 -form  e ~ α  corresponding to  e α {:(13.6)g_(alpha beta)omega^(beta)" is the "1"-form " widetilde(e)_(alpha)" corresponding to "e_(alpha):}\begin{equation*} g_{\alpha \beta} \boldsymbol{\omega}^{\beta} \text { is the } 1 \text {-form } \widetilde{\boldsymbol{e}}_{\alpha} \text { corresponding to } \boldsymbol{e}_{\alpha} \tag{13.6} \end{equation*}(13.6)gαβωβ is the 1-form e~α corresponding to eα
Also as in flat spacetime (§3.2), a tensor can accept either a vector or a 1-form into any given slot
(13.7) S ( u ~ , σ , v ) S ( u , σ , v ) (13.7) S ( u ~ , σ , v ) S ( u , σ , v ) {:(13.7)S( widetilde(u)","sigma","v)-=S(u","sigma","v):}\begin{equation*} \boldsymbol{S}(\widetilde{u}, \sigma, v) \equiv \boldsymbol{S}(u, \sigma, v) \tag{13.7} \end{equation*}(13.7)S(u~,σ,v)S(u,σ,v)
Equivalently, in component language, the indices of a tensor can be lowered with the covariant components of the metric
The basis vectors { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} can be chosen arbitrarily at each event. Therefore the corresponding components g α β g α β g_(alpha beta)g_{\alpha \beta}gαβ of the metric are quite arbitrary (though symmetric: g α β = g β α g α β = g β α g_(alpha beta)=g_(beta alpha)g_{\alpha \beta}=g_{\beta \alpha}gαβ=gβα. But the mixed components g α β g α β g^(alpha)_(beta)g^{\alpha}{ }_{\beta}gαβ are not arbitrary. In particular, equations (13.5) and (13.7) imply
(13.9) g ( u ~ , v ) g ( u , v ) u ~ , v (13.9) g ( u ~ , v ) g ( u , v ) u ~ , v {:(13.9)g( widetilde(u)","v)-=g(u","v)-=(: widetilde(u)","v:):}\begin{equation*} \boldsymbol{g}(\widetilde{\boldsymbol{u}}, \boldsymbol{v}) \equiv \boldsymbol{g}(\boldsymbol{u}, \boldsymbol{v}) \equiv\langle\widetilde{\boldsymbol{u}}, \boldsymbol{v}\rangle \tag{13.9} \end{equation*}(13.9)g(u~,v)g(u,v)u~,v
Therefore one concludes that the metric tensor in mixed representation is identical with the unit matrix:
(13.10) g α β g ( ω α , e β ) ω α , e β = δ α β . (13.10) g α β g ω α , e β ω α , e β = δ α β . {:(13.10)g^(alpha)_(beta)-=g(omega^(alpha),e_(beta))-=(:omega^(alpha),e_(beta):)=delta^(alpha)_(beta).:}\begin{equation*} g^{\alpha}{ }_{\beta} \equiv \boldsymbol{g}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right) \equiv\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle=\delta^{\alpha}{ }_{\beta} . \tag{13.10} \end{equation*}(13.10)gαβg(ωα,eβ)ωα,eβ=δαβ.
This feature of the metric in turn fixes the contravariant components of the metric:
(13.11) g α μ g μ β = g α β = δ α β ; ["lowering an index" of g α μ ] (13.11) g α μ g μ β = g α β = δ α β ;  ["lowering an index" of  g α μ {:[(13.11)g^(alpha mu)g_(mu beta)=g^(alpha)_(beta)=delta^(alpha)_(beta);],[" ["lowering an index" of "{:g^(alpha mu)]]:}\begin{align*} g^{\alpha \mu} g_{\mu \beta} & =g^{\alpha}{ }_{\beta}=\delta^{\alpha}{ }_{\beta} ; \tag{13.11}\\ & \text { ["lowering an index" of } \left.g^{\alpha \mu}\right] \end{align*}(13.11)gαμgμβ=gαβ=δαβ; ["lowering an index" of gαμ]
i.e.,
(13.12) g α β is the matrix inverse of g α β . (13.12) g α β  is the matrix inverse of  g α β {:(13.12)||g^(alpha beta)||" is the matrix inverse of "||g_(alpha beta)||". ":}\begin{equation*} \left\|g^{\alpha \beta}\right\| \text { is the matrix inverse of }\left\|g_{\alpha \beta}\right\| \text {. } \tag{13.12} \end{equation*}(13.12)gαβ is the matrix inverse of gαβ
This reciprocity enables one to undo the lowering of tensor indices (i.e., raise indices) with g α β g α β g^(alpha beta)g^{\alpha \beta}gαβ :
(13.13) S μ β γ = δ μ α S α β γ = g μ ν g ν α S α β γ = g μ ν S ν β γ . (13.13) S μ β γ = δ μ α S α β γ = g μ ν g ν α S α β γ = g μ ν S ν β γ . {:(13.13)S^(mu beta)_(gamma)=delta^(mu)_(alpha)S^(alpha beta)_(gamma)=g^(mu nu)g_(nu alpha)S^(alpha beta)_(gamma)=g^(mu nu)S_(nu)^(beta)_(gamma).:}\begin{equation*} S^{\mu \beta}{ }_{\gamma}=\delta^{\mu}{ }_{\alpha} S^{\alpha \beta}{ }_{\gamma}=g^{\mu \nu} g_{\nu \alpha} S^{\alpha \beta}{ }_{\gamma}=g^{\mu \nu} S_{\nu}{ }^{\beta}{ }_{\gamma} . \tag{13.13} \end{equation*}(13.13)Sμβγ=δμαSαβγ=gμνgναSαβγ=gμνSνβγ.
The last two paragraphs may be summarized in brief:
(1) g α β = δ α β g α β = δ α β g^(alpha)_(beta)=delta^(alpha)_(beta)g^{\alpha}{ }_{\beta}=\delta^{\alpha}{ }_{\beta}gαβ=δαβ;
(2) g α β = g α β 1 g α β = g α β 1 ||g^(alpha beta)||=||g_(alpha beta)||^(-1)\left\|g^{\alpha \beta}\right\|=\left\|g_{\alpha \beta}\right\|^{-1}gαβ=gαβ1;
(3) tensor indices are lowered with g α β g α β g_(alpha beta)g_{\alpha \beta}gαβ;
(4) tensor indices are raised with g α β g α β g^(alpha beta)g^{\alpha \beta}gαβ.
In this formalism of metric and index shuffling, a big question demands attention: how can one tell whether the metric is locally Lorentz rather than locally Euclidean or locally something else? Of course, one criterion (necessary; not sufficient!) is dimensionality-a locally Lorentz spacetime must have four dimensions. (Recall the method of § 1.2 § 1.2 §1.2\S 1.2§1.2 to determine dimensionality.) Confine attention, then, to fourdimensional manifolds. What else must one demand? One must demand that at every event P P P\mathscr{P}P there exist an orthonormal frame (orthonormal set of basis vectors { e α ^ } ) e α ^ {:{e_( hat(alpha))})\left.\left\{\boldsymbol{e}_{\hat{\alpha}}\right\}\right){eα^}) in which the components of the metric have their flat-spacetime form
(13.14) g α ^ β ^ e α ^ e β ^ = η α β diagonal ( 1 , 1 , 1 , 1 ) (13.14) g α ^ β ^ e α ^ e β ^ = η α β  diagonal  ( 1 , 1 , 1 , 1 ) {:(13.14)g_( hat(alpha) hat(beta))-=e_( hat(alpha))*e_( hat(beta))=eta_(alpha beta)-=" diagonal "(-1","1","1","1):}\begin{equation*} g_{\hat{\alpha} \hat{\beta}} \equiv \boldsymbol{e}_{\hat{\alpha}} \cdot \boldsymbol{e}_{\hat{\beta}}=\eta_{\alpha \beta} \equiv \text { diagonal }(-1,1,1,1) \tag{13.14} \end{equation*}(13.14)gα^β^eα^eβ^=ηαβ diagonal (1,1,1,1)
To test for this is straightforward (exercise 13.1). (1) Search for a timelike vector u ( u u < 0 ) u ( u u < 0 ) u(u*u < 0)\boldsymbol{u}(\boldsymbol{u} \cdot \boldsymbol{u}<0)u(uu<0). If none exist, spacetime is not locally Lorentz. If one is found, then (2) examine all non-zero vectors v v v\boldsymbol{v}v perpendicular to u u u\boldsymbol{u}u. If they are all spacelike ( v v > 0 ) ( v v > 0 ) (v*v > 0)(\boldsymbol{v} \cdot \boldsymbol{v}>0)(vv>0), then spacetime is locally Lorentz. Otherwise it is not.

EXERCISES

Exercise 13.1. TEST WHETHER SPACETIME IS LOCAL LORENTZ

Prove that the above two-step procedure for testing whether spacetime is locally Lorentz is valid: i.e., prove that if the procedure says "yes," then there exists an orthonormal basis with g α ^ β ^ = η α β g α ^ β ^ = η α β g_( hat(alpha) hat(beta))=eta_(alpha beta)g_{\hat{\alpha} \hat{\beta}}=\eta_{\alpha \beta}gα^β^=ηαβ at the event in question; if it says "no," then no such basis exists.

Exercise 13.2. PRACTICE WITH METRIC

A four-dimensional manifold with coordinates v , r , θ , ϕ v , r , θ , ϕ v,r,theta,phiv, r, \theta, \phiv,r,θ,ϕ has line element (old-style notation)
d s 2 = ( 1 2 M / r ) d v 2 + 2 d v d r + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = ( 1 2 M / r ) d v 2 + 2 d v d r + r 2 d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-(1-2M//r)dv^(2)+2dvdr+r^(2)(dtheta^(2)+sin^(2)theta dphi^(2))d s^{2}=-(1-2 M / r) d v^{2}+2 d v d r+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)ds2=(12M/r)dv2+2dvdr+r2(dθ2+sin2θdϕ2)
corresponding to metric (new-style notation)
d s 2 = ( 1 2 M / r ) d v d v + d v d r + d r d v + r 2 ( d θ d θ + sin 2 θ d ϕ d ϕ ) d s 2 = ( 1 2 M / r ) d v d v + d v d r + d r d v + r 2 d θ d θ + sin 2 θ d ϕ d ϕ ds^(2)=-(1-2M//r)dv ox dv+dv ox dr+dr ox dv+r^(2)(d theta ox d theta+sin^(2)theta d phi ox d phi)\boldsymbol{d} \boldsymbol{s}^{2}=-(1-2 M / r) \boldsymbol{d} v \otimes \boldsymbol{d} v+\boldsymbol{d} v \otimes \boldsymbol{d} r+\boldsymbol{d} r \otimes \boldsymbol{d} v+r^{2}\left(\boldsymbol{d} \theta \otimes \boldsymbol{d} \theta+\sin ^{2} \theta \boldsymbol{d} \phi \otimes \boldsymbol{d} \phi\right)ds2=(12M/r)dvdv+dvdr+drdv+r2(dθdθ+sin2θdϕdϕ)
where M M MMM is a constant.
(a) Find the "covariant" components g α β g α β g_(alpha beta)g_{\alpha \beta}gαβ and "contravariant" components g α β g α β g^(alpha beta)g^{\alpha \beta}gαβ of the metric in this coordinate system. [Answer: g v v = ( 1 2 M / r ) , g v r = g r v = 1 , g θ θ = r 2 , g ϕ ϕ = r 2 g v v = ( 1 2 M / r ) , g v r = g r v = 1 , g θ θ = r 2 , g ϕ ϕ = r 2 g_(vv)=-(1-2M//r),g_(vr)=g_(rv)=1,g_(theta theta)=r^(2),g_(phi phi)=r^(2)g_{v v}=-(1-2 M / r), g_{v r}=g_{r v}=1, g_{\theta \theta}=r^{2}, g_{\phi \phi}=r^{2}gvv=(12M/r),gvr=grv=1,gθθ=r2,gϕϕ=r2 sin 2 θ sin 2 θ sin^(2)theta\sin ^{2} \thetasin2θ; all other g α β g α β g_(alpha beta)g_{\alpha \beta}gαβ vanish; g v r = g r v = 1 , g r r = ( 1 2 M / r ) , g θ θ = r 2 , g ϕ ϕ = r 2 sin 2 θ g v r = g r v = 1 , g r r = ( 1 2 M / r ) , g θ θ = r 2 , g ϕ ϕ = r 2 sin 2 θ g^(vr)=g^(rv)=1,g^(rr)=(1-2M//r),g^(theta theta)=r^(-2),g^(phi phi)=r^(-2)sin^(-2)thetag^{v r}=g^{r v}=1, g^{r r}=(1-2 M / r), g^{\theta \theta}=r^{-2}, g^{\phi \phi}=r^{-2} \sin ^{-2} \thetagvr=grv=1,grr=(12M/r),gθθ=r2,gϕϕ=r2sin2θ, all other g α β g α β g^(alpha beta)g^{\alpha \beta}gαβ vanish.]
(b) Define a scalar field t t ttt by
t v r 2 M ln [ ( r / 2 M ) 1 ] t v r 2 M ln [ ( r / 2 M ) 1 ] t-=v-r-2M ln[(r//2M)-1]t \equiv v-r-2 M \ln [(r / 2 M)-1]tvr2Mln[(r/2M)1]
What are the covariant and contravariant components ( u α u α u_(alpha)u_{\alpha}uα and u α u α u^(alpha)u^{\alpha}uα ) of the 1 -form u ~ d t u ~ d t widetilde(u)-=dt\widetilde{\boldsymbol{u}} \equiv \boldsymbol{d} tu~dt ? What is the squared length u 2 u u u 2 u u u^(2)-=u*u\boldsymbol{u}^{2} \equiv \boldsymbol{u} \cdot \boldsymbol{u}u2uu, of the corresponding vector? Show that u u u\boldsymbol{u}u is timelike in the region r > 2 M r > 2 M r > 2Mr>2 Mr>2M. [Answer: u v = 1 , u r = 1 / ( 1 2 M / r ) , u θ = u ϕ = 0 ; u v = 1 / ( 1 u v = 1 , u r = 1 / ( 1 2 M / r ) , u θ = u ϕ = 0 ; u v = 1 / ( 1 u_(v)=1,u_(r)=-1//(1-2M//r),u_(theta)=u_(phi)=0;u^(v)=-1//(1-u_{v}=1, u_{r}=-1 /(1-2 M / r), u_{\theta}=u_{\phi}=0 ; u^{v}=-1 /(1-uv=1,ur=1/(12M/r),uθ=uϕ=0;uv=1/(1 2 M / r ) , u r = 0 , u θ = u ϕ = 0 ; u 2 = 1 / ( 1 2 M / r ) 2 M / r ) , u r = 0 , u θ = u ϕ = 0 ; u 2 = 1 / ( 1 2 M / r ) 2M//r),u^(r)=0,u^(theta)=u^(phi)=0;u^(2)=-1//(1-2M//r)2 M / r), u^{r}=0, u^{\theta}=u^{\phi}=0 ; \boldsymbol{u}^{2}=-1 /(1-2 M / r)2M/r),ur=0,uθ=uϕ=0;u2=1/(12M/r).]
(c) Find the most general non-zero vector w w w\boldsymbol{w}w orthogonal to u u u\boldsymbol{u}u in the region r > 2 M r > 2 M r > 2Mr>2 Mr>2M, and show that it is spacelike. Thereby conclude that spacetime is locally Lorentz in the region r > 2 M r > 2 M r > 2Mr>2 Mr>2M. [Answer: Since w u = w α u α = w v / ( 1 2 M / r ) , w v w u = w α u α = w v / ( 1 2 M / r ) , w v w*u=w_(alpha)u^(alpha)=-w_(v)//(1-2M//r),w_(v)\boldsymbol{w} \cdot \boldsymbol{u}=w_{\alpha} u^{\alpha}=-w_{v} /(1-2 M / r), w_{v}wu=wαuα=wv/(12M/r),wv must vanish, but w r , w θ , w ϕ w r , w θ , w ϕ w_(r),w_(theta),w_(phi)w_{r}, w_{\theta}, w_{\phi}wr,wθ,wϕ are arbitrary, and w 2 = ( 1 2 M / r ) w r 2 + r 2 w θ 2 + r 2 sin 2 θ w ϕ 2 > 0 w 2 = ( 1 2 M / r ) w r 2 + r 2 w θ 2 + r 2 sin 2 θ w ϕ 2 > 0 w^(2)=(1-2M//r)w_(r)^(2)+r^(-2)w_(theta)^(2)+r^(-2)sin^(-2)thetaw_(phi)^(2) > 0\boldsymbol{w}^{2}=(1-2 M / r) w_{r}{ }^{2}+r^{-2} w_{\theta}{ }^{2}+r^{-2} \sin ^{-2} \theta w_{\phi}^{2}>0w2=(12M/r)wr2+r2wθ2+r2sin2θwϕ2>0.]
(d) Let t , r , θ , ϕ t , r , θ , ϕ t,r,theta,phit, r, \theta, \phit,r,θ,ϕ be new coordinates for spacetime. Find the line element in this coordinate system. [Answer: This is the "Schwarzschild" line element
d s 2 = ( 1 2 M r ) d t 2 + d r 2 1 2 M / r + r 2 d θ 2 + r 2 sin 2 θ d ϕ 2 . ] d s 2 = 1 2 M r d t 2 + d r 2 1 2 M / r + r 2 d θ 2 + r 2 sin 2 θ d ϕ 2 . {:ds^(2)=-(1-(2M)/(r))dt^(2)+(dr^(2))/(1-2M//r)+r^(2)dtheta^(2)+r^(2)sin^(2)theta dphi^(2).]\left.d s^{2}=-\left(1-\frac{2 M}{r}\right) d t^{2}+\frac{d r^{2}}{1-2 M / r}+r^{2} d \theta^{2}+r^{2} \sin ^{2} \theta d \phi^{2} .\right]ds2=(12Mr)dt2+dr212M/r+r2dθ2+r2sin2θdϕ2.]
(e) Find an orthonormal basis, for which g α ^ β ^ = η α β g α ^ β ^ = η α β g_( hat(alpha) hat(beta))=eta_(alpha beta)g_{\hat{\alpha} \hat{\beta}}=\eta_{\alpha \beta}gα^β^=ηαβ in the region r > 2 M r > 2 M r > 2Mr>2 Mr>2M. [Answer: e 0 ^ ( 1 2 M / r ) 1 / 2 / t , e r ^ ( 1 2 M / r ) 1 / 2 / r , e θ ^ r 1 / θ , e ϕ ^ = ( r sin θ ) 1 / ϕ . ] e 0 ^ ( 1 2 M / r ) 1 / 2 / t , e r ^ ( 1 2 M / r ) 1 / 2 / r , e θ ^ r 1 / θ , e ϕ ^ = ( r sin θ ) 1 / ϕ . {:e_( hat(0))-=(1-2M//r)^(-1//2)del//del t,e_( hat(r))-=(1-2M//r)^(1//2)del//del r,e_( hat(theta))-=r^(-1)del//del theta,e_( hat(phi))=(r sin theta)^(-1)del//del phi.]\left.\boldsymbol{e}_{\hat{0}} \equiv(1-2 M / r)^{-1 / 2} \partial / \partial t, \boldsymbol{e}_{\hat{r}} \equiv(1-2 M / r)^{1 / 2} \partial / \partial r, \boldsymbol{e}_{\hat{\theta}} \equiv r^{-1} \partial / \partial \theta, \boldsymbol{e}_{\hat{\phi}}=(r \sin \theta)^{-1} \partial / \partial \phi.\right]e0^(12M/r)1/2/t,er^(12M/r)1/2/r,eθ^r1/θ,eϕ^=(rsinθ)1/ϕ.]

§13.3. CONCORD BETWEEN GEODESICS OF CURVED SPACETIME GEOMETRY AND STRAIGHT LINES OF LOCAL LORENTZ GEOMETRY

More could be said about the mathematical machinery and physical implications of "metric," but an issue of greater urgency presses for attention. What has metric (or spacetime interval) to do with geodesic (or world line of test particle)? Answer:
Two mathematical objects ("straight line in a local Lorentz frame" and "geodesic of the over-all global curved spacetime geometry") equal to the same physical object ("world line of test particle") must be equal to each other ("condition of consistency"). As a first method to spell out this consistency requirement, examine the two mathematical representations of the world line of a test particle in the neighborhood of a given event P 0 P 0 P_(0)\mathscr{P}_{0}P0. The local-Lorentz representation says:
"Pick a local Lorentz frame at P 0 P 0 P_(0)\mathscr{P}_{0}P0. [As spelled out in exercise 13.3, such a local Lorentz frame is the closest thing there is to a global Lorentz frame at P 0 P 0 P_(0)\mathscr{P}_{0}P0; i.e., it is a coordinate system in which
(13.15a) g α β ( P 0 ) = η α β ( flat-spacetime metric ) , (13.15b) g α β , γ ( P 0 ) = 0 , (13.15c) g α β , γ δ ( P 0 ) 0 except in special cases, such as flat space.] (13.15a) g α β P 0 = η α β (  flat-spacetime metric  ) , (13.15b) g α β , γ P 0 = 0 , (13.15c) g α β , γ δ P 0 0  except in special cases, such as flat space.]  {:[(13.15a)g_(alpha beta)(P_(0))=eta_(alpha beta)(" flat-spacetime metric ")","],[(13.15b)g_(alpha beta,gamma)(P_(0))=0","],[(13.15c)g_(alpha beta,gamma delta)(P_(0))!=0" except in special cases, such as flat space.] "]:}\begin{align*} g_{\alpha \beta}\left(\mathscr{P}_{0}\right) & =\eta_{\alpha \beta}(\text { flat-spacetime metric }), \tag{13.15a}\\ g_{\alpha \beta, \gamma}\left(\mathscr{P}_{0}\right) & =0, \tag{13.15b}\\ g_{\alpha \beta, \gamma \delta}\left(\mathscr{P}_{0}\right) & \neq 0 \text { except in special cases, such as flat space.] } \tag{13.15c} \end{align*}(13.15a)gαβ(P0)=ηαβ( flat-spacetime metric ),(13.15b)gαβ,γ(P0)=0,(13.15c)gαβ,γδ(P0)0 except in special cases, such as flat space.] 
The world line in that frame has zero acceleration,
(13.16) d 2 x α / d τ 2 = 0 at P 0 ("straight-line equation"), (13.16) d 2 x α / d τ 2 = 0  at  P 0  ("straight-line equation"),  {:(13.16)d^(2)x^(alpha)//dtau^(2)=0" at "P_(0)" ("straight-line equation"), ":}\begin{equation*} d^{2} x^{\alpha} / d \tau^{2}=0 \text { at } \mathscr{P}_{0} \text { ("straight-line equation"), } \tag{13.16} \end{equation*}(13.16)d2xα/dτ2=0 at P0 ("straight-line equation"), 
where τ τ tau\tauτ is proper time as measured by the particle's clock."
The geodesic representation says
"In the local Lorentz frame, as in any coordinate frame, the world line satisfies the geodesic equation
(13.17) d 2 x α / d τ 2 + Γ α β γ ( d x β / d τ ) ( d x γ / d τ ) = 0 (13.17) d 2 x α / d τ 2 + Γ α β γ d x β / d τ d x γ / d τ = 0 {:(13.17)d^(2)x^(alpha)//dtau^(2)+Gamma^(alpha)_(beta gamma)(dx^(beta)//d tau)(dx^(gamma)//d tau)=0:}\begin{equation*} d^{2} x^{\alpha} / d \tau^{2}+\Gamma^{\alpha}{ }_{\beta \gamma}\left(d x^{\beta} / d \tau\right)\left(d x^{\gamma} / d \tau\right)=0 \tag{13.17} \end{equation*}(13.17)d2xα/dτ2+Γαβγ(dxβ/dτ)(dxγ/dτ)=0
( τ τ tau\tauτ is an affine parameter because it is time as measured by the test particle's clock)." Consistency of the two representations for any and every choice of test particle (any and every choice of d x α / d τ d x α / d τ dx^(alpha)//d taud x^{\alpha} / d \taudxα/dτ at P 0 P 0 P_(0)\mathscr{P}_{0}P0 ) demands
(13.18) Γ β γ α ( P 0 ) = 0 in any local Lorentz frame [coordinate system satisfying equations (13.15) at P 0 ] ; (13.18) Γ β γ α P 0 = 0  in any local Lorentz frame [coordinate   system satisfying equations (13.15) at  P 0 ; {:[(13.18)Gamma_(beta gamma)^(alpha)(P_(0))=0],[" in any local Lorentz frame [coordinate "],[" system satisfying equations (13.15) at "{:P_(0)];]:}\begin{align*} & \Gamma_{\beta \gamma}^{\alpha}\left(\mathscr{P}_{0}\right)= 0 \tag{13.18}\\ & \text { in any local Lorentz frame [coordinate } \\ &\text { system satisfying equations (13.15) at } \left.\mathscr{P}_{0}\right] ; \end{align*}(13.18)Γβγα(P0)=0 in any local Lorentz frame [coordinate  system satisfying equations (13.15) at P0];
i.e., it demands that every local Lorentz frame is a local inertial frame. (On local inertial frames see §11.6.) In such a frame, all local effects of "gravitation" disappear. That is the physical shorthand for (13.18).
One does not have to speak in the language of a specific coordinate system when one demands identity between the geodesic (derived from the Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ ) and the straight line of the local Lorentz geometry ( g μ ν ) g μ ν (g_(mu nu))\left(g_{\mu \nu}\right)(gμν). The local Lorentz specialization of coordinates may be the most immediate way to see the physics ("no local effects of gravitation"), but it is not the right way to formulate the basic mathematical requirement in its full generality and power. The right way is to demand
(13.19) g = 0 ("compatibility of g and "). (13.19) g = 0  ("compatibility of  g  and   ").  {:(13.19)grad g=0" ("compatibility of "g" and "grad" "). ":}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{g}=0 \text { ("compatibility of } \boldsymbol{g} \text { and } \boldsymbol{\nabla} \text { "). } \tag{13.19} \end{equation*}(13.19)g=0 ("compatibility of g and  "). 
Stated in the language of an arbitrary coordinate system, this requirement reads
(13.19') g α β ; γ g α β x γ Γ α γ μ g μ β Γ μ β γ g α μ = 0 . (13.19') g α β ; γ g α β x γ Γ α γ μ g μ β Γ μ β γ g α μ = 0 . {:(13.19')g_(alpha beta;gamma)-=(delg_(alpha beta))/(delx^(gamma))-Gamma_(alpha gamma)^(mu)g_(mu beta)-Gamma^(mu)_(beta gamma)g_(alpha mu)=0.:}\begin{equation*} g_{\alpha \beta ; \gamma} \equiv \frac{\partial g_{\alpha \beta}}{\partial x^{\gamma}}-\Gamma_{\alpha \gamma}^{\mu} g_{\mu \beta}-\Gamma^{\mu}{ }_{\beta \gamma} g_{\alpha \mu}=0 . \tag{13.19'} \end{equation*}(13.19')gαβ;γgαβxγΓαγμgμβΓμβγgαμ=0.
Local-Lorentz description of straight lines
Geodesic description of straight lines
Condition of consistency: Γ α β γ = 0 Γ α β γ = 0 Gamma^(alpha)_(beta gamma)=0\Gamma^{\alpha}{ }_{\beta \gamma}=0Γαβγ=0 in local Lorentz frame
Consistency reformulated:
g = 0 g = 0 grad g=0\boldsymbol{\nabla} \boldsymbol{g}=0g=0.
That this covariant requirement is fulfilled in every coordinate system follows from its validity in one coordinate system: a local Lorentz frame. (The first term in this equation, and the last two terms, are separately required to vanish in the local Lorentz frame at point P 0 P 0 P_(0)\mathscr{P}_{0}P0-and required to vanish by the physics.) From g = 0 g = 0 grad g=0\boldsymbol{\nabla} \boldsymbol{g}=0g=0, one can derive both the abstract chain rule
(13.20) u ( v w ) = ( u v ) w + v ( u w ) (13.20) u ( v w ) = u v w + v u w {:(13.20)grad_(u)(v*w)=(grad_(u)v)*w+v*(grad_(u)w):}\begin{equation*} \nabla_{u}(v \cdot w)=\left(\nabla_{u} v\right) \cdot w+v \cdot\left(\nabla_{u} w\right) \tag{13.20} \end{equation*}(13.20)u(vw)=(uv)w+v(uw)
Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ expressed in terms of metric
(Exercise 13.4) and the following equations for the connection coefficients in any frame in terms of (1) the metric coefficients, g α β = e α e β g α β = e α e β g_(alpha beta)=e_(alpha)*e_(beta)g_{\alpha \beta}=\boldsymbol{e}_{\alpha} \cdot \boldsymbol{e}_{\beta}gαβ=eαeβ, and (2) the covariant commutation coefficients
(13.21) c α β γ c α β μ g μ γ ω μ , [ e α , e β ] g μ γ (13.21) c α β γ c α β μ g μ γ ω μ , e α , e β g μ γ {:(13.21)c_(alpha beta gamma)-=c_(alpha beta)^(mu)g_(mu gamma)-=(:omega^(mu),[e_(alpha),e_(beta)]:)g_(mu gamma):}\begin{equation*} c_{\alpha \beta \gamma} \equiv c_{\alpha \beta}{ }^{\mu} g_{\mu \gamma} \equiv\left\langle\boldsymbol{\omega}^{\mu},\left[\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right]\right\rangle g_{\mu \gamma} \tag{13.21} \end{equation*}(13.21)cαβγcαβμgμγωμ,[eα,eβ]gμγ
of that frame:
(13.22) Γ β γ α = g α μ Γ μ β γ (definition of Γ μ β γ ) , Γ μ β γ = 1 2 ( g μ β , γ + c μ β γ + g μ γ , β + c μ γ β g β γ , μ c β γ μ ) (13.23) = 1 2 ( g μ β , γ + g μ γ , β g β γ , μ ) in any coordinate frame. (13.22) Γ β γ α = g α μ Γ μ β γ  (definition of  Γ μ β γ , Γ μ β γ = 1 2 g μ β , γ + c μ β γ + g μ γ , β + c μ γ β g β γ , μ c β γ μ (13.23) = 1 2 g μ β , γ + g μ γ , β g β γ , μ  in any coordinate frame.  {:[(13.22){:Gamma_(beta gamma)^(alpha)=g^(alpha mu)Gamma_(mu beta gamma)quad" (definition of "Gamma_(mu beta gamma))","],[Gamma_(mu beta gamma)=(1)/(2)(g_(mu beta,gamma)+c_(mu beta gamma)+g_(mu gamma,beta)+c_(mu gamma beta)-g_(beta gamma,mu)-c_(beta gamma mu))],[(13.23)=(1)/(2)(g_(mu beta,gamma)+g_(mu gamma,beta)-g_(beta gamma,mu))" in any coordinate frame. "]:}\begin{gather*} \left.\Gamma_{\beta \gamma}^{\alpha}=g^{\alpha \mu} \Gamma_{\mu \beta \gamma} \quad \text { (definition of } \Gamma_{\mu \beta \gamma}\right), \tag{13.22}\\ \Gamma_{\mu \beta \gamma}=\frac{1}{2}\left(g_{\mu \beta, \gamma}+c_{\mu \beta \gamma}+g_{\mu \gamma, \beta}+c_{\mu \gamma \beta}-g_{\beta \gamma, \mu}-c_{\beta \gamma \mu}\right) \\ =\frac{1}{2}\left(g_{\mu \beta, \gamma}+g_{\mu \gamma, \beta}-g_{\beta \gamma, \mu}\right) \text { in any coordinate frame. } \tag{13.23} \end{gather*}(13.22)Γβγα=gαμΓμβγ (definition of Γμβγ),Γμβγ=12(gμβ,γ+cμβγ+gμγ,β+cμγβgβγ,μcβγμ)(13.23)=12(gμβ,γ+gμγ,βgβγ,μ) in any coordinate frame. 
(See Exercise 13.4).
Equations (13.23) are the connection coefficients required to make the geodesics of curved spacetime coincide with the straight lines of the local Lorentz geometry. And they are fixed uniquely; no other choice of connection coefficients will do the job!
Summary: in curved spacetime with a local Lorentz metric, the following seemingly different statements are actually equivalent: (1) the geodesics of curved spacetime coincide with the straight lines of the local Lorentz geometry; (2) every local Lorentz frame [coordinates with g α β ( P 0 ) = η α β , g α β , γ ( P 0 ) = 0 g α β P 0 = η α β , g α β , γ P 0 = 0 g_(alpha beta)(P_(0))=eta_(alpha beta),g_(alpha beta,gamma)(P_(0))=0g_{\alpha \beta}\left(\mathscr{P}_{0}\right)=\eta_{\alpha \beta}, g_{\alpha \beta, \gamma}\left(\mathscr{P}_{0}\right)=0gαβ(P0)=ηαβ,gαβ,γ(P0)=0 ] is a local inertial frame [ Γ α β γ ( P 0 ) = 0 ] Γ α β γ P 0 = 0 [Gamma^(alpha)_(beta gamma)(P_(0))=0]\left[\Gamma^{\alpha}{ }_{\beta \gamma}\left(\mathscr{P}_{0}\right)=0\right][Γαβγ(P0)=0]; (3) the metric and covariant derivative satisfy the compatibility condition g = 0 g = 0 grad g=0\boldsymbol{\nabla} \boldsymbol{g}=0g=0; (4) the covariant derivative obeys the chain rule (13.20); (5) the connection coefficients are determined by the metric in the manner of equations (13.23). A sixth equivalent statement, derived in the next section, says (6) the geodesics of curved spacetime coincide with world lines of extremal proper time.

EXERCISES

Exercise 13.3. MATHEMATICAL REPRESENTATION OF LOCAL

LORENTZ FRAME
By definition, a local Lorentz frame at a given event P 0 P 0 P_(0)\mathscr{P}_{0}P0 is the closest thing there to a global Lorentz frame. Thus, it should be a coordinate system with g μ ν ( P 0 ) = η μ ν g μ ν P 0 = η μ ν g_(mu nu)(P_(0))=eta_(mu nu)g_{\mu \nu}\left(\mathscr{P}_{0}\right)=\eta_{\mu \nu}gμν(P0)=ημν, and with as many derivatives of g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν as possible vanishing at P 0 P 0 P_(0)\mathscr{\mathscr { P }}_{0}P0. Prove that there exist coordinates in which g μ ν ( P 0 ) = η μ ν g μ ν P 0 = η μ ν g_(mu nu)(P_(0))=eta_(mu nu)g_{\mu \nu}\left(\mathscr{P}_{0}\right)=\eta_{\mu \nu}gμν(P0)=ημν and g μ ν , o ( P 0 ) = 0 g μ ν , o P 0 = 0 g_(mu nu,o)(P_(0))=0g_{\mu \nu, o}\left(\mathscr{P}_{0}\right)=0gμν,o(P0)=0, but that g μ ν , o σ ( P 0 ) g μ ν , o σ P 0 g_(mu nu,o sigma)(P_(0))g_{\mu \nu, o \sigma}\left(\mathscr{P}_{0}\right)gμν,oσ(P0) cannot vanish in general. Hence, such coordinates are the mathematical representation of a local Lorentz frame. [Hint: Let { x α ( P ) } x α ( P ) {x^({:alpha^(')(P)}):}\left\{x^{\left.\alpha^{\prime}(\mathscr{P})\right\}}\right.{xα(P)} be an arbitrary but specific coordinate system, and { x μ ( P ) } x μ ( P ) {x^(mu)(P)}\left\{x^{\mu}(\mathscr{P})\right\}{xμ(P)} be a local Lorentz frame, both
with origins at P 0 P 0 P_(0)\mathscr{P}_{0}P0. Expand the coordinate transformation between the two in powers of x μ x μ x^(mu)x^{\mu}xμ
x α = M α μ x μ + 1 2 N α μ ν x μ x ν + 1 6 P α μ ν ρ x μ x ν x ρ + ; x α = M α μ x μ + 1 2 N α μ ν x μ x ν + 1 6 P α μ ν ρ x μ x ν x ρ + ; x^(alpha^('))=M^(alpha)_(mu)x^(mu)+(1)/(2)N^(alpha)_(mu nu)x^(mu)x^(nu)+(1)/(6)P^(alpha)_(mu nu rho)x^(mu)x^(nu)x^(rho)+dots;x^{\alpha^{\prime}}=M^{\alpha}{ }_{\mu} x^{\mu}+\frac{1}{2} N^{\alpha}{ }_{\mu \nu} x^{\mu} x^{\nu}+\frac{1}{6} P^{\alpha}{ }_{\mu \nu \rho} x^{\mu} x^{\nu} x^{\rho}+\ldots ;xα=Mαμxμ+12Nαμνxμxν+16Pαμνρxμxνxρ+;
and use the transformation matrix L α μ x α / x μ L α μ x α / x μ L^(alpha^('))_(mu)-=delx^(alpha^('))//delx^(mu)L^{\alpha^{\prime}}{ }_{\mu} \equiv \partial x^{\alpha^{\prime}} / \partial x^{\mu}Lαμxα/xμ to get g μ ν ( P 0 ) , g μ ν , ρ ( P 0 ) g μ ν P 0 , g μ ν , ρ P 0 g_(mu nu)(P_(0)),g_(mu nu,rho)(P_(0))g_{\mu \nu}\left(\mathscr{P}_{0}\right), g_{\mu \nu, \rho}\left(\mathscr{P}_{0}\right)gμν(P0),gμν,ρ(P0), and g μ ν , ρ σ ( P 0 ) g μ ν , ρ σ P 0 g_(mu nu,rho sigma)(P_(0))g_{\mu \nu, \rho \sigma}\left(\mathscr{P}_{0}\right)gμν,ρσ(P0) in terms of g α β g α β g_(alpha^(')beta^('))g_{\alpha^{\prime} \beta^{\prime}}gαβ and its derivatives and the constants M α μ , N α μ ν , P α μ ν ρ M α μ , N α μ ν , P α μ ν ρ M^(alpha)_(mu),N^(alpha)_(mu nu),P^(alpha)_(mu nurho^('))M^{\alpha}{ }_{\mu}, N^{\alpha}{ }_{\mu \nu}, P^{\alpha}{ }_{\mu \nu \rho^{\prime}}Mαμ,Nαμν,Pαμνρ. Show that whatever g α β g α β g_(alpha^(')beta^('))g_{\alpha^{\prime} \beta^{\prime}}gαβ may be (so long as it is nonsingular, so g α β g α β g^(alpha^(')beta^('))g^{\alpha^{\prime} \beta^{\prime}}gαβ exists!), one can choose the 16 constants M α μ M α μ M^(alpha)_(mu)M^{\alpha}{ }_{\mu}Mαμ to make g μ ν = η μ ν g μ ν = η μ ν g_(mu nu)=eta_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}gμν=ημν (ten conditions); one can choose the 4 × 10 = 40 4 × 10 = 40 4xx10=404 \times 10=404×10=40 constants N γ μ ν N γ μ ν N^(gamma)_(mu nu)N^{\gamma}{ }_{\mu \nu}Nγμν to make the 10 × 4 = 40 g μ ν , ρ ( P 0 ) 10 × 4 = 40 g μ ν , ρ P 0 10 xx4=40g_(mu nu,rho)(P_(0))10 \times 4=40 g_{\mu \nu, \rho}\left(\mathscr{P}_{0}\right)10×4=40gμν,ρ(P0) vanish; but one cannot in general choose the 4 × 20 = 4 × 20 = 4xx20=4 \times 20=4×20= 80 P α μ ν ρ 80 P α μ ν ρ 80P^(alpha)_(mu nu rho)80 P^{\alpha}{ }_{\mu \nu \rho}80Pαμνρ to make the 10 × 10 = 100 g μ ν , ρ σ 10 × 10 = 100 g μ ν , ρ σ 10 xx10=100g_(mu nu,rho sigma)10 \times 10=100 g_{\mu \nu, \rho \sigma}10×10=100gμν,ρσ vanish.]

Exercise 13.4. CONSEQUENCES OF COMPATIBILITY BETWEEN g g g\boldsymbol{g}g AND grad\boldsymbol{\nabla}

(a) From the condition of compatibility g = 0 g = 0 grad g=0\boldsymbol{\nabla} \boldsymbol{g}=0g=0, derive the chain rule (13.20).
(b) From the condition of compatibility g = 0 g = 0 grad g=0\boldsymbol{\nabla} \boldsymbol{g}=0g=0 and definitions (13.21) and (13.22), derive equation (13.23) for the connection coefficients. [Answer: See exercise 8.15, p. 216.]

§13.4. GEODESICS AS WORLD LINES OF EXTREMAL PROPER TIME

In a local Lorentz frame, it is easy to distinguish a world line that is straight from one that is not. Position the Lorentz frame and so orient it that the starting point of the world line, a a aaa, lies at the origin and the end point, B B B\mathscr{B}B, lies at x = 0 , y = 0 x = 0 , y = 0 x=0,y=0x=0, y=0x=0,y=0, z = 0 , t = T z = 0 , t = T z=0,t=Tz=0, t=Tz=0,t=T. As an example of a nonstraight world line, consider passage at uniform velocity from a a a\mathscr{a}a to point P P P\mathscr{P}P with coordinates ( 1 2 T ; 0 , 0 , 1 2 R ) 1 2 T ; 0 , 0 , 1 2 R ((1)/(2)T;0,0,(1)/(2)R)\left(\frac{1}{2} T ; 0,0, \frac{1}{2} R\right)(12T;0,0,12R) and from there again with uniform velocity to point B B B\mathscr{B}B. The lapse of proper time from start to finish ("length of world line") is
τ = ( T 2 R 2 ) 1 / 2 τ = T 2 R 2 1 / 2 tau=(T^(2)-R^(2))^(1//2)\tau=\left(T^{2}-R^{2}\right)^{1 / 2}τ=(T2R2)1/2
Thus the lapse of proper time is diminished from its straight-line value, and diminished moreover for any choice of R R RRR whatsoever, except for the zero or straight-line value R = 0 R = 0 R=0R=0R=0. As for this simple nonstraight curve, so also for any other nonstraight curve: the lapse of proper time between A A A\mathscr{A}A and G G G\mathscr{G}G is less than the straight-line lapse (Exercise 6.3). Thus, in flat spacetime, extremal length of world line is an indicator of straightness.
Any local region of the curved spacetime of the real, physical world is Lorentz in character. In this local Lorentz geometry, it is easy to set up Lorentz coordinates and carry out the extremal-length analysis just sketched to distinguish between a straight line and a nonstraight line:
τ = a B d τ = a B ( η μ ν d x μ d x ν ) 1 / 2 (13.24) = ( a maximum for straight line as compared to any variant of the straight line ) . τ = a B d τ = a B η μ ν d x μ d x ν 1 / 2 (13.24) =  a maximum for straight line   as compared to any variant of   the straight line  . {:[tau=int_(a)^(B)d tau=int_(a)^(B)(-eta_(mu nu)dx^(mu)dx^(nu))^(1//2)],[(13.24)=([" a maximum for straight line "],[" as compared to any variant of "],[" the straight line "]).]:}\begin{align*} \tau & =\int_{a}^{\mathscr{B}} d \tau=\int_{a}^{\mathscr{B}}\left(-\eta_{\mu \nu} d x^{\mu} d x^{\nu}\right)^{1 / 2} \\ & =\left(\begin{array}{l} \text { a maximum for straight line } \\ \text { as compared to any variant of } \\ \text { the straight line } \end{array}\right) . \tag{13.24} \end{align*}τ=aBdτ=aB(ημνdxμdxν)1/2(13.24)=( a maximum for straight line  as compared to any variant of  the straight line ).
In flat spacetime, straight lines have extremal length
Extremal length in curved spacetime
Such a test for straightness can be carried out separately in each local Lorentz region along the world line, or, with greater efficiency, it can be carried out over many local Lorentz regions simultaneously, i.e., over a region with endpoints C C C\mathscr{C}C and B B B\mathscr{B}B so widely separated that no single Lorentz frame can possibly contain them both. To carry out the analysis, one must abandon local Lorentz coordinates. Therefore introduce a general curvilinear coordinate system and find
τ = Q B d τ = a B ( g μ ν d x μ d x ν ) 1 / 2 (13.25) = ( an extremum for timelike world line that is straight in each local Lorentz frame along its path, as compared to any "nearby" variant of this world line ) . τ = Q B d τ = a B g μ ν d x μ d x ν 1 / 2 (13.25) =  an extremum for timelike world line that   is straight in each local Lorentz frame   along its path, as compared to any "nearby"   variant of this world line  . {:[tau=int_(Q)^(B)d tau=int_(a)^(B)(-g_(mu nu)dx^(mu)dx^(nu))^(1//2)],[(13.25)=([" an extremum for timelike world line that "],[" is straight in each local Lorentz frame "],[" along its path, as compared to any "nearby" "],[" variant of this world line "]).]:}\begin{align*} \tau & =\int_{\mathscr{Q}}^{\mathscr{B}} d \tau=\int_{a}^{\mathscr{B}}\left(-g_{\mu \nu} d x^{\mu} d x^{\nu}\right)^{1 / 2} \\ & =\left(\begin{array}{l} \text { an extremum for timelike world line that } \\ \text { is straight in each local Lorentz frame } \\ \text { along its path, as compared to any "nearby" } \\ \text { variant of this world line } \end{array}\right) . \tag{13.25} \end{align*}τ=QBdτ=aB(gμνdxμdxν)1/2(13.25)=( an extremum for timelike world line that  is straight in each local Lorentz frame  along its path, as compared to any "nearby"  variant of this world line ).
In the real world, the path of extremal τ τ tau\tauτ, being straight in every local Lorentz frame, must be a geodesic of spacetime.
Notice that the word "maximum" in equation (13.24) has been replaced by "extremum" in the statement (13.25). When a a aaa and B B B\mathscr{B}B are widely separated, they may be connected by several different geodesics with differing lapses of proper time (Figure 13.2). Each timelike geodesic extremizes τ τ tau\tauτ with respect to nearby deformations of itself, but the extremum need not be a maximum. When several distinct geodesics connect two events, the typical one is not a local maximum ("mountain peak") but a saddle point ("mountain pass") in such a diagram as Figure 13.2 or 13.3.
Concord between locally straight lines (lines of extremal τ τ tau\tauτ ) and geodesics of curved spacetime demands that timelike geodesics have extremal proper length. If so, then any curve x μ ( λ ) x μ ( λ ) x^(mu)(lambda)x^{\mu}(\lambda)xμ(λ) between a a a\mathscr{a}a (where λ = 0 λ = 0 lambda=0\lambda=0λ=0 ) and B B B\mathscr{B}B (where λ = 1 λ = 1 lambda=1\lambda=1λ=1 ) that extremizes τ τ tau\tauτ should satisfy the geodesic equation. To test for an extremal by comparing times, pick a curve suspected to be a geodesic, and deform it slightly but arbitrarily:
original curve, x μ = a μ ( λ ) (13.26) deformed curve, x μ = a μ ( λ ) + δ a μ ( λ ) .  original curve,  x μ = a μ ( λ ) (13.26)  deformed curve,  x μ = a μ ( λ ) + δ a μ ( λ ) . {:[" original curve, "x^(mu)=a^(mu)(lambda)],[(13.26)" deformed curve, "x^(mu)=a^(mu)(lambda)+deltaa^(mu)(lambda).]:}\begin{align*} \text { original curve, } x^{\mu} & =a^{\mu}(\lambda) \\ \text { deformed curve, } x^{\mu} & =a^{\mu}(\lambda)+\delta a^{\mu}(\lambda) . \tag{13.26} \end{align*} original curve, xμ=aμ(λ)(13.26) deformed curve, xμ=aμ(λ)+δaμ(λ).
Along either curve the lapse of proper time is
(13.27) τ = a F d τ = 0 1 ( g μ ν d x μ d λ d x ν d λ ) 1 / 2 d λ (13.27) τ = a F d τ = 0 1 g μ ν d x μ d λ d x ν d λ 1 / 2 d λ {:(13.27)tau=int_(a)^(F)d tau=int_(0)^(1)(-g_(mu nu)(dx^(mu))/(d lambda)(dx^(nu))/(d lambda))^(1//2)d lambda:}\begin{equation*} \tau=\int_{a}^{\mathscr{F}} d \tau=\int_{0}^{1}\left(-g_{\mu \nu} \frac{d x^{\mu}}{d \lambda} \frac{d x^{\nu}}{d \lambda}\right)^{1 / 2} d \lambda \tag{13.27} \end{equation*}(13.27)τ=aFdτ=01(gμνdxμdλdxνdλ)1/2dλ
At fixed λ λ lambda\lambdaλ the metric coefficient g μ ν [ x α ( λ ) ] g μ ν x α ( λ ) g_(mu nu)[x^(alpha)(lambda)]g_{\mu \nu}\left[x^{\alpha}(\lambda)\right]gμν[xα(λ)] differs from one curve to the other by
(13.28) δ g μ ν g μ ν [ a α ( λ ) + δ a α ( λ ) ] g μ ν [ a α ( λ ) ] = g μ ν x σ δ a σ ( λ ) (13.28) δ g μ ν g μ ν a α ( λ ) + δ a α ( λ ) g μ ν a α ( λ ) = g μ ν x σ δ a σ ( λ ) {:(13.28)deltag_(mu nu)-=g_(mu nu)[a^(alpha)(lambda)+deltaa^(alpha)(lambda)]-g_(mu nu)[a^(alpha)(lambda)]=(delg_(mu nu))/(delx^(sigma))deltaa^(sigma)(lambda):}\begin{equation*} \delta g_{\mu \nu} \equiv g_{\mu \nu}\left[a^{\alpha}(\lambda)+\delta a^{\alpha}(\lambda)\right]-g_{\mu \nu}\left[a^{\alpha}(\lambda)\right]=\frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} \delta a^{\sigma}(\lambda) \tag{13.28} \end{equation*}(13.28)δgμνgμν[aα(λ)+δaα(λ)]gμν[aα(λ)]=gμνxσδaσ(λ)
and the components d x ν / d λ d x ν / d λ dx^(nu)//d lambdad x^{\nu} / d \lambdadxν/dλ of the tangent vector differ by
(13.29) δ ( d x ν d λ ) d ( a ν + δ a ν ) d λ d a ν d λ = d d λ ( δ a ν ) (13.29) δ d x ν d λ d a ν + δ a ν d λ d a ν d λ = d d λ δ a ν {:(13.29)delta((dx^(nu))/(d lambda))-=(d(a^(nu)+deltaa^(nu)))/(d lambda)-(da^(nu))/(d lambda)=(d)/(d lambda)(deltaa^(nu)):}\begin{equation*} \delta\left(\frac{d x^{\nu}}{d \lambda}\right) \equiv \frac{d\left(a^{\nu}+\delta a^{\nu}\right)}{d \lambda}-\frac{d a^{\nu}}{d \lambda}=\frac{d}{d \lambda}\left(\delta a^{\nu}\right) \tag{13.29} \end{equation*}(13.29)δ(dxνdλ)d(aν+δaν)dλdaνdλ=ddλ(δaν)
These changes in g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν and d x ν / d λ d x ν / d λ dx^(nu)//d lambdad x^{\nu} / d \lambdadxν/dλ, at fixed λ λ lambda\lambdaλ, produce corresponding changes in the lapse of proper time in equation (13.27):
δ τ = 0 1 { g μ ν ( d a μ / d λ ) d ( δ a ν ) / d λ 1 2 ( g μ ν , σ δ a σ ) ( d a μ / d λ ) ( d a ν / d λ ) [ g γ δ ( d a γ / d λ ) ( d a δ / d λ ) ] 1 / 2 } d λ . δ τ = 0 1 g μ ν d a μ / d λ d δ a ν / d λ 1 2 g μ ν , σ δ a σ d a μ / d λ d a ν / d λ g γ δ d a γ / d λ d a δ / d λ 1 / 2 d λ . delta tau=int_(0)^(1){(-g_(mu nu)(da^(mu)//d lambda)d(deltaa^(nu))//d lambda-(1)/(2)(g_(mu nu,sigma)deltaa^(sigma))(da^(mu)//d lambda)(da^(nu)//d lambda))/([-g_(gamma delta)(da^(gamma)//d lambda)(da^(delta)//d lambda)]^(1//2))}d lambda.\delta \tau=\int_{0}^{1}\left\{\frac{-g_{\mu \nu}\left(d a^{\mu} / d \lambda\right) d\left(\delta a^{\nu}\right) / d \lambda-\frac{1}{2}\left(g_{\mu \nu, \sigma} \delta a^{\sigma}\right)\left(d a^{\mu} / d \lambda\right)\left(d a^{\nu} / d \lambda\right)}{\left[-g_{\gamma \delta}\left(d a^{\gamma} / d \lambda\right)\left(d a^{\delta} / d \lambda\right)\right]^{1 / 2}}\right\} d \lambda .δτ=01{gμν(daμ/dλ)d(δaν)/dλ12(gμν,σδaσ)(daμ/dλ)(daν/dλ)[gγδ(daγ/dλ)(daδ/dλ)]1/2}dλ.
Integrate the first term by parts. Strike out the end-point terms, because both paths must pass through a a aaa and B ( δ a μ = 0 B δ a μ = 0 B(deltaa^(mu)=0:}\mathscr{B}\left(\delta a^{\mu}=0\right.B(δaμ=0 at λ = 0 λ = 0 lambda=0\lambda=0λ=0 and λ = 1 ) λ = 1 {: lambda=1)\left.\lambda=1\right)λ=1). Thus find
(13.30) δ τ = λ = 0 λ = 1 f σ ( λ ) δ a σ [ g γ δ d a γ d λ d a δ d λ ] 1 / 2 d λ . (13.30) δ τ = λ = 0 λ = 1 f σ ( λ ) δ a σ g γ δ d a γ d λ d a δ d λ 1 / 2 d λ . {:(13.30)delta tau=int_(lambda=0)^(lambda=1)f_(sigma)(lambda)deltaa^(sigma)[-g_(gamma delta)(da^(gamma))/(d lambda)(da^(delta))/(d lambda)]^(1//2)d lambda.:}\begin{equation*} \delta \tau=\int_{\lambda=0}^{\lambda=1} f_{\sigma}(\lambda) \delta a^{\sigma}\left[-g_{\gamma \delta} \frac{d a^{\gamma}}{d \lambda} \frac{d a^{\delta}}{d \lambda}\right]^{1 / 2} d \lambda . \tag{13.30} \end{equation*}(13.30)δτ=λ=0λ=1fσ(λ)δaσ[gγδdaγdλdaδdλ]1/2dλ.
Here the f σ f σ f_(sigma)f_{\sigma}fσ ("force terms") in the integrand are abbreviations for the four expressions
f σ ( λ ) = 1 [ g γ δ d a γ d λ d a δ d λ ] 1 / 2 d d λ g σ ν d a ν d λ [ g γ δ d a γ d λ d a δ d λ ] 1 / 2 1 2 g μ ν x σ d a μ d λ d a ν d λ [ g γ δ d a γ d λ d a δ d λ ] f σ ( λ ) = 1 g γ δ d a γ d λ d a δ d λ 1 / 2 d d λ g σ ν d a ν d λ g γ δ d a γ d λ d a δ d λ 1 / 2 1 2 g μ ν x σ d a μ d λ d a ν d λ g γ δ d a γ d λ d a δ d λ f_(sigma)(lambda)=(1)/([-g_(gamma delta)(da^(gamma))/(d lambda)(da^(delta))/(d lambda)]^(1//2))(d)/(d lambda)(g_(sigma nu)(da^(nu))/(d lambda))/([-g_(gamma delta)(da^(gamma))/(d lambda)(da^(delta))/(d lambda)]^(1//2))-((1)/(2)(delg_(mu nu))/(delx^(sigma))(da^(mu))/(d lambda)(da^(nu))/(d lambda))/([-g_(gamma delta)(da^(gamma))/(d lambda)(da^(delta))/(d lambda)])f_{\sigma}(\lambda)=\frac{1}{\left[-g_{\gamma \delta} \frac{d a^{\gamma}}{d \lambda} \frac{d a^{\delta}}{d \lambda}\right]^{1 / 2}} \frac{d}{d \lambda} \frac{g_{\sigma \nu} \frac{d a^{\nu}}{d \lambda}}{\left[-g_{\gamma \delta} \frac{d a^{\gamma}}{d \lambda} \frac{d a^{\delta}}{d \lambda}\right]^{1 / 2}}-\frac{\frac{1}{2} \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} \frac{d a^{\mu}}{d \lambda} \frac{d a^{\nu}}{d \lambda}}{\left[-g_{\gamma \delta} \frac{d a^{\gamma}}{d \lambda} \frac{d a^{\delta}}{d \lambda}\right]}fσ(λ)=1[gγδdaγdλdaδdλ]1/2ddλgσνdaνdλ[gγδdaγdλdaδdλ]1/212gμνxσdaμdλdaνdλ[gγδdaγdλdaδdλ].
An extremum is achieved, and the first-order change δ τ δ τ delta tau\delta \tauδτ vanishes for every first-order deformation δ a σ ( λ ) δ a σ ( λ ) deltaa^(sigma)(lambda)\delta a^{\sigma}(\lambda)δaσ(λ) from an optimal path x σ = a σ ( λ ) x σ = a σ ( λ ) x^(sigma)=a^(sigma)(lambda)x^{\sigma}=a^{\sigma}(\lambda)xσ=aσ(λ), when the four quantities f σ f σ f_(sigma)f_{\sigma}fσ that multiply the δ a σ δ a σ deltaa^(sigma)\delta a^{\sigma}δaσ all vanish. Thus one arrives at the four conditions
(13.32) f σ ( λ ) = 0 (13.32) f σ ( λ ) = 0 {:(13.32)f_(sigma)(lambda)=0:}\begin{equation*} f_{\sigma}(\lambda)=0 \tag{13.32} \end{equation*}(13.32)fσ(λ)=0
for the determination of an extremal world line. (An alternative viewpoint on the extremization is spelled out in Figure 13.3.)
Sufficient these four equations are, but independent they are not, by reason of a "bead argument" (automatic vanishing of δ τ δ τ delta tau\delta \tauδτ for any set of changes that merely slide points, like beads, along an existing world line). The operation of mere "sliding of beads" implies the trivial change
(13.33) δ a σ ( λ ) = h ( λ ) d a σ d λ (13.33) δ a σ ( λ ) = h ( λ ) d a σ d λ {:(13.33)deltaa^(sigma)(lambda)=h(lambda)(da^(sigma))/(d lambda):}\begin{equation*} \delta a^{\sigma}(\lambda)=h(\lambda) \frac{d a^{\sigma}}{d \lambda} \tag{13.33} \end{equation*}(13.33)δaσ(λ)=h(λ)daσdλ
where h ( λ ) h ( λ ) h(lambda)h(\lambda)h(λ) is an arbitrary function of position along the world line ("more sliding here than there"). Already knowing that this operation cannot change τ τ tau\tauτ, one is guaranteed that the integrand in (13.30) must vanish when one inserts (13.33) for δ a σ δ a σ deltaa^(sigma)\delta a^{\sigma}δaσ; and must vanish, moreover, whatever choice is made for the arbitrary "magnitude of slide" factor h ( λ ) h ( λ ) h(lambda)h(\lambda)h(λ). This requirement implies and demands that the scalar product f σ d a σ / d λ f σ d a σ / d λ f_(sigma)da^(sigma)//d lambdaf_{\sigma} d a^{\sigma} / d \lambdafσdaσ/dλ must automatically vanish; or, otherwise stated,
(13.34) f σ d a σ d τ = 0 . (13.34) f σ d a σ d τ = 0 . {:(13.34)f_(sigma)(da^(sigma))/(d tau)=0.:}\begin{equation*} f_{\sigma} \frac{d a^{\sigma}}{d \tau}=0 . \tag{13.34} \end{equation*}(13.34)fσdaσdτ=0.
The argument applies, and this equation holds, whether one is or is not dealing with an optimal world line. An equation of this type, valid whether or not the world line is an allowable track for a free test particle (track of extremal lapse of proper
time), is known as an identity. Equation (13.34), an important identity in the realm of spacetime geodesics, is an appropriate forerunner for the Bianchi identities of Chapter 15: the most important identities in the realm of spacetime curvature.
The freedom that exists to "slide λ λ lambda\lambdaλ-values along the world line" can be exploited to replace the arbitrary parameter λ λ lambda\lambdaλ by the physically more interesting parameter of proper time itself,
(13.35) d τ = [ g γ δ d a γ d λ d a δ d λ ] 1 / 2 d λ (13.35) d τ = g γ δ d a γ d λ d a δ d λ 1 / 2 d λ {:(13.35)d tau=[-g_(gamma delta)(da^(gamma))/(d lambda)(da^(delta))/(d lambda)]^(1//2)d lambda:}\begin{equation*} d \tau=\left[-g_{\gamma \delta} \frac{d a^{\gamma}}{d \lambda} \frac{d a^{\delta}}{d \lambda}\right]^{1 / 2} d \lambda \tag{13.35} \end{equation*}(13.35)dτ=[gγδdaγdλdaδdλ]1/2dλ
Figure 13.2.
Star oscillating back and forth through the plane of a disc galaxy, as an example of a situation where two events A A A\mathscr{A}A and B B B\mathscr{B}B can be connected by more than one geodesic. Upper left: The galaxy seen edge-on, showing (dashed line) the path of the star in question, referred to a local frame partaking of and comoving with the general revolution of the nearby "dise stars." Upper right: The effective potential sensed by the star, according to Newtonian gravitation theory, is like that experienced by a ball which rolls down one inclined plane and up another ("free fall toward galactic plane" with acceleration g = 1 2 g = 1 2 g=(1)/(2)g=\frac{1}{2}g=12 in the units used here). The three central frames: Possible and impossible world lines for the star connecting two given events A A A\mathscr{A}A (plane of galaxy at t = 0 t = 0 t=0t=0t=0 ) and B B B\mathscr{B}B (plane of galaxy at t = 2 t = 2 t=2t=2t=2 ). Right: Throw star up from the galactic plane with enough velocity so that it just gets back to the plane at t = 2 t = 2 t=2t=2t=2. Left: Throw it up with half the velocity and it will come back in half the time (very contrary to behavior of a simple harmonic oscillation, but in accord with galaxy's v-shaped potential!), thus being able to make two excursions in the allotted time between A A A\mathscr{A}A and B B B\mathscr{B}B. Center: A conceivable world line (conceivable with rocket propulsion!) but not a geodesic. Bottom: Comparison of these and any other paths that allow themselves to be approximated in the form
z = a 1 sin ( π t / 2 ) + a 2 sin ( 2 π t / 2 ) . z = a 1 sin ( π t / 2 ) + a 2 sin ( 2 π t / 2 ) . z=a_(1)sin(pi t//2)+a_(2)sin(2pi t//2).z=a_{1} \sin (\pi t / 2)+a_{2} \sin (2 \pi t / 2) .z=a1sin(πt/2)+a2sin(2πt/2).
Here the two adjustable parameters, a 1 a 1 a_(1)a_{1}a1 and a 2 a 2 a_(2)a_{2}a2, provide the coordinates in a two-dimensional "function space" (approximation to the infinite-dimensional function space required to depict all conceivable world lines connecting G G G\mathscr{G}G and B B B\mathscr{B}B; note comparison in right center frame between one-term Fourier approximation and exact, parabolic law of free fall; similarly in left center frame, where the two curves agree too closely to be shown separate on the diagram). Details: In the context of general relativity, take an arbitrary world line that connects C C C\mathscr{C}C and B B B\mathscr{B}B, evaluate lapse of proper time, repeat for other world lines, and say that a given world line represents a possible motion ("geodesic") when for it the proper time is an extremum with respect to all nearby world lines. In the Newtonian approximation, the difference between the lapse of proper time and the lapse ( t S B t d t S B t d t_(SB)-t_(d)t_{S B}-t_{d}tSBtd ) of coordinate time is all that comes to attention, in the form of the "action integral" (on a "per-unit-mass basis")
I = a S [ ( kinetic energy ) ( potential energy ) ] d t = [ 1 2 ( d z d t ) 2 | z | ] d t I = a S (  kinetic   energy  ) (  potential   energy  ) d t = 1 2 d z d t 2 | z | d t {:[I=int_(a)^(S)[((" kinetic ")/(" energy "))-((" potential ")/(" energy "))]dt],[=int[(1)/(2)((dz)/(dt))^(2)-|z|]dt]:}\begin{aligned} I & =\int_{a}^{S}\left[\binom{\text { kinetic }}{\text { energy }}-\binom{\text { potential }}{\text { energy }}\right] d t \\ & =\int\left[\frac{1}{2}\left(\frac{d z}{d t}\right)^{2}-|z|\right] d t \end{aligned}I=aS[( kinetic  energy )( potential  energy )]dt=[12(dzdt)2|z|]dt
(maximum, or other extremum, in the proper time implies minimum, or corresponding other extremum, in the action I I III ). The integration gives
I = ( π 2 a 1 2 / 8 ) ( 4 | a 1 | / π ) + ( π 2 a 2 2 / 2 ) I = π 2 a 1 2 / 8 4 a 1 / π + π 2 a 2 2 / 2 I=(pi^(2)a_(1)^(2)//8)-(4|a_(1)|//pi)+(pi^(2)a_(2)^(2)//2)I=\left(\pi^{2} a_{1}^{2} / 8\right)-\left(4\left|a_{1}\right| / \pi\right)+\left(\pi^{2} a_{2}^{2} / 2\right)I=(π2a12/8)(4|a1|/π)+(π2a22/2)
for | a 2 | < 1 2 | a 1 | a 2 < 1 2 a 1 |a_(2)| < (1)/(2)|a_(1)|\left|a_{2}\right|<\frac{1}{2}\left|a_{1}\right||a2|<12|a1| (one-excursion motions), and for | a 2 | > 1 2 | a 1 | a 2 > 1 2 a 1 |a_(2)| > (1)/(2)|a_(1)|\left|a_{2}\right|>\frac{1}{2}\left|a_{1}\right||a2|>12|a1| (two-excursion motions),
I = ( π 2 a 1 2 / 8 ) + ( π 2 a 2 2 / 2 ) ( 4 | a 2 | / π ) ( a 1 2 / π | a 2 | ) I = π 2 a 1 2 / 8 + π 2 a 2 2 / 2 4 a 2 / π a 1 2 / π a 2 I=(pi^(2)a_(1)^(2)//8)+(pi^(2)a_(2)^(2)//2)-(4|a_(2)|//pi)-(a_(1)^(2)//pi|a_(2)|)I=\left(\pi^{2} a_{1}^{2} / 8\right)+\left(\pi^{2} a_{2}^{2} / 2\right)-\left(4\left|a_{2}\right| / \pi\right)-\left(a_{1}^{2} / \pi\left|a_{2}\right|\right)I=(π2a12/8)+(π2a22/2)(4|a2|/π)(a12/π|a2|)
The one-excursion motion minimizes the action (maximizes the lapse of proper time). The two-excursion motion extremizes the action but does not minimize it ("saddle point"; "mountain pass" in the topography). Choquard (1955) gives other examples of problems of mechanics where there is more than one extremum. Morse (1934) and Morse and Cairns (1969) give a theorem connecting the number of saddles of various types with the numbers of maxima and minima ("critical-point theorem of the calculus of variations in the large").

Figure 13.3.
Extremizing lapse of proper time by suitable choice of world line. Left: Spacetime; and world line F F FFF that extremizes the lapse of proper time τ τ tau\tauτ from a a aaa to B B B\mathscr{B}B compared to other world lines. The specific world lines depicted in the diagram happen to be distinguished from fiducial world line G G GGG by two "Fourier amplitudes" a 1 a 1 a_(1)a_{1}a1 and a 2 a 2 a_(2)a_{2}a2 :
δ a μ ( λ ) = a 1 sin ( π λ ) + a 2 sin ( 2 π λ ) δ a μ ( λ ) = a 1 sin ( π λ ) + a 2 sin ( 2 π λ ) deltaa^(mu)(lambda)=a_(1)sin(pi lambda)+a_(2)sin(2pi lambda)\delta a^{\mu}(\lambda)=a_{1} \sin (\pi \lambda)+a_{2} \sin (2 \pi \lambda)δaμ(λ)=a1sin(πλ)+a2sin(2πλ)
where the arbitrary scaling of λ λ lambda\lambdaλ, and its zero, are so adjusted that λ ( C ) = 0 , λ ( B ) = 1 λ ( C ) = 0 , λ ( B ) = 1 lambda(C)=0,lambda(B)=1\lambda(\mathscr{C})=0, \lambda(\mathscr{B})=1λ(C)=0,λ(B)=1.
Right: "Path space." The coordinates in this space are the Fourier amplitudes a 1 a 1 a_(1)a_{1}a1 and a 2 a 2 a_(2)a_{2}a2. Only these two amplitudes ("two dimensions") are shown out of what in principle are infinitely many amplitudes ("infinite-dimensional path space") required to represent the general timelike world line connecting a a aaa and D D D\mathscr{\mathcal { D }}D. Any given contour curve runs through all those points (in path space) for which the corresponding world lines (in spacetime) rack up the indicated lapse of proper time τ τ tau\tauτ. Foregoing description is classical; according to quantum mechanics, all the timelike world lines connecting C C C\mathscr{C}C and B B B\mathscr{B}B occur with the same probability amplitude ("principle of democracy of histories") with the only difference from one to another being the phase of this complex probability amplitude exp ( i m τ / ) ( m = exp ( i m τ / ) ( m = exp(-im tau//ℏ)(m=\exp (-i m \tau / \hbar)(m=exp(imτ/)(m= mass of particle, = = ℏ=\hbar== quantum of angular momentum). In the sum over these probability amplitudes, however, destructive interference wipes out the contributions from all those histories which differ too much from the optimal or classical history ("Fresnel wave zone"; "Feynman's principle of sum over histories"; see Feynman and Hibbs, 1965). Capitalizing on this wave-mechanical background to show how the machinery of the physical world works, Box 25.3 spells out the Hamilton-Jacobi method ("short-wavelength limit of quantum mechanics") for determining geodesics, a method considerably more convenient for most applications than the usual "second-order differential equations for geodesics" (equation 10.27).
Focus on a specific world line, x μ = a μ ( λ ) x μ = a μ ( λ ) x^(mu)=a^(mu)(lambda)x^{\mu}=a^{\mu}(\lambda)xμ=aμ(λ), with all deformations of it gone from view; one may replace a μ ( λ ) a μ ( λ ) a^(mu)(lambda)a^{\mu}(\lambda)aμ(λ) by x μ ( λ ) x μ ( λ ) x^(mu)(lambda)x^{\mu}(\lambda)xμ(λ) everywhere. Then the differential equations (13.32) for an extremal world line reduce to
(13.36) g σ ν d 2 x ν d τ 2 + 1 2 ( g σ ν x μ + g σ μ x ν g μ ν x σ ) d x μ d τ d x ν d τ = 0 . (13.36) g σ ν d 2 x ν d τ 2 + 1 2 g σ ν x μ + g σ μ x ν g μ ν x σ d x μ d τ d x ν d τ = 0 . {:(13.36)g_(sigma nu)(d^(2)x^(nu))/(dtau^(2))+(1)/(2)((delg_(sigma nu))/(delx^(mu))+(delg_(sigma mu))/(delx^(nu))-(delg_(mu nu))/(delx^(sigma)))(dx^(mu))/(d tau)(dx^(nu))/(d tau)=0.:}\begin{equation*} g_{\sigma \nu} \frac{d^{2} x^{\nu}}{d \tau^{2}}+\frac{1}{2}\left(\frac{\partial g_{\sigma \nu}}{\partial x^{\mu}}+\frac{\partial g_{\sigma \mu}}{\partial x^{\nu}}-\frac{\partial g_{\mu \nu}}{\partial x^{\sigma}}\right) \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu}}{d \tau}=0 . \tag{13.36} \end{equation*}(13.36)gσνd2xνdτ2+12(gσνxμ+gσμxνgμνxσ)dxμdτdxνdτ=0.
As an aside, note that the identity (13.34) now follows by one differentiation (with respect to τ τ tau\tauτ ) of the equation
(13.37) g σ ν d x σ d τ d x ν d τ + 1 = 0 (13.37) g σ ν d x σ d τ d x ν d τ + 1 = 0 {:(13.37)g_(sigma nu)(dx^(sigma))/(d tau)(dx^(nu))/(d tau)+1=0:}\begin{equation*} g_{\sigma \nu} \frac{d x^{\sigma}}{d \tau} \frac{d x^{\nu}}{d \tau}+1=0 \tag{13.37} \end{equation*}(13.37)gσνdxσdτdxνdτ+1=0
Thus the identity is to be interpreted as saying that 4 -velocity and 4 -acceleration are orthogonal for any world line, extremal or not. Now return to (13.36), raise an index with g β σ g β σ g^(beta sigma)g^{\beta \sigma}gβσ, and thereby bring the equation for a straight line of local Lorentz geometry into the form
(13.38) d 2 x β d τ 2 + g β σ 1 2 ( g σ ν x μ + g σ μ x ν g μ ν x σ ) d x μ d τ d x ν d τ = 0 (13.38) d 2 x β d τ 2 + g β σ 1 2 g σ ν x μ + g σ μ x ν g μ ν x σ d x μ d τ d x ν d τ = 0 {:(13.38)(d^(2)x^(beta))/(dtau^(2))+g^(beta sigma)(1)/(2)((delg_(sigma nu))/(delx^(mu))+(delg_(sigma mu))/(delx^(nu))-(delg_(mu nu))/(delx^(sigma)))(dx^(mu))/(d tau)(dx^(nu))/(d tau)=0:}\begin{equation*} \frac{d^{2} x^{\beta}}{d \tau^{2}}+g^{\beta \sigma} \frac{1}{2}\left(\frac{\partial g_{\sigma \nu}}{\partial x^{\mu}}+\frac{\partial g_{\sigma \mu}}{\partial x^{\nu}}-\frac{\partial g_{\mu \nu}}{\partial x^{\sigma}}\right) \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu}}{d \tau}=0 \tag{13.38} \end{equation*}(13.38)d2xβdτ2+gβσ12(gσνxμ+gσμxνgμνxσ)dxμdτdxνdτ=0
Compare with the standard form of the equation for a geodesic in "premetric geometry,"
(13.39) d 2 x β d λ 2 + Γ β μ ν d x μ d λ d x ν d λ = 0 (13.39) d 2 x β d λ 2 + Γ β μ ν d x μ d λ d x ν d λ = 0 {:(13.39)(d^(2)x^(beta))/(dlambda^(2))+Gamma^(beta)_(mu nu)(dx^(mu))/(d lambda)(dx^(nu))/(d lambda)=0:}\begin{equation*} \frac{d^{2} x^{\beta}}{d \lambda^{2}}+\Gamma^{\beta}{ }_{\mu \nu} \frac{d x^{\mu}}{d \lambda} \frac{d x^{\nu}}{d \lambda}=0 \tag{13.39} \end{equation*}(13.39)d2xβdλ2+Γβμνdxμdλdxνdλ=0
Conclude that the geodesics of the premetric geometry will agree with the straight lines of the local Lorentz geometry if and only if two conditions are satisfied: (1) the 40 connection coefficients Γ β μ ν Γ β μ ν Gamma^(beta)_(mu nu)\Gamma^{\beta}{ }_{\mu \nu}Γβμν that define geodesics, covariant derivatives, and parallel transport must be given in terms of the 10 metric coefficients g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν ("Einstein gravitation potentials") by the equations (13.22) and (13.23) previously derived; and (2) the geodesic parameter λ λ lambda\lambdaλ must agree with the proper time τ τ tau\tauτ up to an arbitrary normalization of zero point and an arbitrary but constant scale factor; thus
λ = a τ + b . λ = a τ + b . lambda=a tau+b.\lambda=a \tau+b .λ=aτ+b.
(Nothing in the formalism has any resemblance whatsoever to the universal time t t ttt of Newton "flowing everywhere uniformly"; rather, there is a separate proper time τ τ tau\tauτ for each geodesic). See Box 13.3 for another variational principle, which gives in one step both the extremal world line and the right parametrization on that line.
With this step, one has completed the transfer of the ideas of curved-space geometry from a foundation based on geodesics to a foundation based on metric. The resulting geometry always and everywhere anchors itself to the principle of "local Lorentz character," as the geometry of Newton-Cartan never did and never could.

Exercise 13.5. ONCE TIMELIKE, ALWAYS TIMELIKE

EXERCISES

Show that a geodesic of spacetime which is timelike at one event is everywhere timelike. Similarly, show that a geodesic initially spacelike is everywhere spacelike, and a geodesic initially null is everywhere null. [Hint: This is the easiest exercise in the book!]

Box 13.2 "GEODESIC" VERSUS "EXTREMAL WORLD LINE"

Once the connection coefficients Γ α μ ν Γ α μ ν Gamma^(alpha)_(mu nu)\Gamma^{\alpha}{ }_{\mu \nu}Γαμν have been expressed in terms of Einstein's gravitational potentials g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν by the equations (13.22) and (13.23), as they are now and hereafter will be in this book ("Riemannian or metric geometry"), it is permissible and appropriate to subsume under the one word "geodesic" two previously distinct ideas: (1) a parametrized world line that satisfies the geodesic equation
d 2 x α d λ 2 + Γ α μ ν d x μ d λ d x ν d λ = 0 ; d 2 x α d λ 2 + Γ α μ ν d x μ d λ d x ν d λ = 0 ; (d^(2)x^(alpha))/(dlambda^(2))+Gamma^(alpha)_(mu nu)(dx^(mu))/(d lambda)(dx^(nu))/(d lambda)=0;\frac{d^{2} x^{\alpha}}{d \lambda^{2}}+\Gamma^{\alpha}{ }_{\mu \nu} \frac{d x^{\mu}}{d \lambda} \frac{d x^{\nu}}{d \lambda}=0 ;d2xαdλ2+Γαμνdxμdλdxνdλ=0;
and (2) a world line that extremizes the proper time (or, if spacelike, a curve that extremizes the proper distance) between two events G G G\mathscr{G}G and B B B\mathscr{B}B. The one possible source of confusion is this, that (1)
presupposes a properly parametrized curve (as was essential, for example, in the Schild's ladder construction employed for parallel transport in Chapter 10 ), whereas (2) cares only about the course of the world line through spacetime, being indifferent to what parametrization is used or whether any parametrization at all is introduced. This is not to deny the possibility of "marking in afterward" along the extremal curve the most natural and easily evaluated of all parameters, the proper time itself, whereupon the extremal curve of (2) satisfies the geodesic equation of (1). Ambiguity is avoided by insisting on proper parametrization: henceforth the word "curve" means a parametrized curve, the word "geodesic" means a properly parametrized geodesic.

Box 13.3 "DYNAMIC" VARIATIONAL PRINCIPLE FOR GEODESICS

If the principle of extremal length
(1) τ = a B [ g μ ν d x μ d λ d x ν d λ ] 1 / 2 d λ = extremum (1) τ = a B g μ ν d x μ d λ d x ν d λ 1 / 2 d λ =  extremum  {:(1)tau=int_(a)^(B)[-g_(mu nu)(dx^(mu))/(d lambda)(dx^(nu))/(d lambda)]^(1//2)d lambda=" extremum ":}\begin{equation*} \tau=\int_{a}^{\mathscr{B}}\left[-g_{\mu \nu} \frac{d x^{\mu}}{d \lambda} \frac{d x^{\nu}}{d \lambda}\right]^{1 / 2} d \lambda=\text { extremum } \tag{1} \end{equation*}(1)τ=aB[gμνdxμdλdxνdλ]1/2dλ= extremum 
is indifferent to choice of parametrization [" d λ d λ d lambdad \lambdadλ " canceling out in (1)] and if the geodesic equation finds the proper parametrization a matter of concern, it is appropriate to search for another extremal principle that yields in one package both the right curve and the right parameter. By analogy with elementary mechanics, one expects that an equation of motion [the geodesic equation
d 2 x μ / d λ 2 + Γ α β μ ( d x α / d λ ) ( d x β / d λ ) = 0 ] d 2 x μ / d λ 2 + Γ α β μ d x α / d λ d x β / d λ = 0 {:d^(2)x^(mu)//dlambda^(2)+Gamma_(alpha beta)^(mu)(dx^(alpha)//d lambda)(dx^(beta)//d lambda)=0]\left.d^{2} x^{\mu} / d \lambda^{2}+\Gamma_{\alpha \beta}^{\mu}\left(d x^{\alpha} / d \lambda\right)\left(d x^{\beta} / d \lambda\right)=0\right]d2xμ/dλ2+Γαβμ(dxα/dλ)(dxβ/dλ)=0]
whose leading term has the form " x ¨ x ¨ x^(¨)\ddot{x}x¨ " can be derived from a Lagrangian with leading term " 1 2 x ˙ 1 2 x ˙ (1)/(2)x^(˙)\frac{1}{2} \dot{x}12x˙ "" ("kinetic energy"; "dynamic" term). The simplest coordinate invariant generalization of 1 2 x ˙ 2 1 2 x ˙ 2 (1)/(2)x^(˙)^(2)\frac{1}{2} \dot{x}^{2}12x˙2 is
1 2 g μ ν ( d x μ / d λ ) ( d x ν / d λ ) 1 2 g μ ν d x μ / d λ d x ν / d λ (1)/(2)g_(mu nu)(dx^(mu)//d lambda)(dx^(nu)//d lambda)\frac{1}{2} g_{\mu \nu}\left(d x^{\mu} / d \lambda\right)\left(d x^{\nu} / d \lambda\right)12gμν(dxμ/dλ)(dxν/dλ)
Thus one is led to try, in place of the "geometric" principle of extremal length, a new "dynamic" extremal principle:
I = 1 2 a B g μ ν d x μ d λ d x ν d λ d λ (2) = a B L ( x σ , d x σ d λ ) d λ = extremum I = 1 2 a B g μ ν d x μ d λ d x ν d λ d λ (2) = a B L x σ , d x σ d λ d λ =  extremum  {:[I=(1)/(2)int_(a)^(B)g_(mu nu)(dx^(mu))/(d lambda)(dx^(nu))/(d lambda)d lambda],[(2)=int_(a)^(B)L(x^(sigma),(dx^(sigma))/(d lambda))d lambda=" extremum "]:}\begin{align*} I & =\frac{1}{2} \int_{a}^{\mathscr{B}} g_{\mu \nu} \frac{d x^{\mu}}{d \lambda} \frac{d x^{\nu}}{d \lambda} d \lambda \\ & =\int_{a}^{\mathscr{B}} L\left(x^{\sigma}, \frac{d x^{\sigma}}{d \lambda}\right) d \lambda=\text { extremum } \tag{2} \end{align*}I=12aBgμνdxμdλdxνdλdλ(2)=aBL(xσ,dxσdλ)dλ= extremum 
(replacement of square root in previous variational principle by first power). The condition for an extremum, here as before [equations (13.30) to (13.32)] is annulment of the so-called Euler-Lagrange "functional derivative"
0 = δ I δ x σ ( coefficient of δ x σ in the integrand of δ I ) (3) = L x σ d d λ L ( d x σ d λ ) 0 = δ I δ x σ (  coefficient of  δ x σ  in   the integrand of  δ I ) (3) = L x σ d d λ L d x σ d λ {:[0=(delta I)/(deltax^(sigma))-=((" coefficient of "deltax^(sigma)" in ")/(" the integrand of "delta I))],[(3)=(del L)/(delx^(sigma))-(d)/(d lambda)(del L)/(del((dx^(sigma))/(d lambda)))]:}\begin{align*} 0 & =\frac{\delta I}{\delta x^{\sigma}} \equiv\binom{\text { coefficient of } \delta x^{\sigma} \text { in }}{\text { the integrand of } \delta I} \\ & =\frac{\partial L}{\partial x^{\sigma}}-\frac{d}{d \lambda} \frac{\partial L}{\partial\left(\frac{d x^{\sigma}}{d \lambda}\right)} \tag{3} \end{align*}0=δIδxσ( coefficient of δxσ in  the integrand of δI)(3)=LxσddλL(dxσdλ)
or, written out in full detail,
(4) g σ ν d 2 x ν d λ 2 + 1 2 ( g σ ν x μ + g σ μ x ν g μ ν x σ ) d x μ d λ d x ν d λ = 0 ; (4) g σ ν d 2 x ν d λ 2 + 1 2 g σ ν x μ + g σ μ x ν g μ ν x σ d x μ d λ d x ν d λ = 0 ; {:(4)g_(sigma nu)(d^(2)x^(nu))/(dlambda^(2))+(1)/(2)((delg_(sigma nu))/(delx^(mu))+(delg_(sigma mu))/(delx^(nu))-(delg_(mu nu))/(delx^(sigma)))(dx^(mu))/(d lambda)(dx^(nu))/(d lambda)=0;:}\begin{equation*} g_{\sigma \nu} \frac{d^{2} x^{\nu}}{d \lambda^{2}}+\frac{1}{2}\left(\frac{\partial g_{\sigma \nu}}{\partial x^{\mu}}+\frac{\partial g_{\sigma \mu}}{\partial x^{\nu}}-\frac{\partial g_{\mu \nu}}{\partial x^{\sigma}}\right) \frac{d x^{\mu}}{d \lambda} \frac{d x^{\nu}}{d \lambda}=0 ; \tag{4} \end{equation*}(4)gσνd2xνdλ2+12(gσνxμ+gσμxνgμνxσ)dxμdλdxνdλ=0;
or, after multiplication by the reciprocal metric,
(5) d 2 x α d λ 2 + g α σ 1 2 ( g o ν x μ + g o μ x ν g μ ν x σ ) d x μ d λ d x ν d λ = 0 (5) d 2 x α d λ 2 + g α σ 1 2 g o ν x μ + g o μ x ν g μ ν x σ d x μ d λ d x ν d λ = 0 {:(5)(d^(2)x^(alpha))/(dlambda^(2))+g^(alpha sigma)(1)/(2)((delg_(o nu))/(delx^(mu))+(delg_(o mu))/(delx^(nu))-(delg_(mu nu))/(delx^(sigma)))(dx^(mu))/(d lambda)(dx^(nu))/(d lambda)=0:}\begin{equation*} \frac{d^{2} x^{\alpha}}{d \lambda^{2}}+g^{\alpha \sigma} \frac{1}{2}\left(\frac{\partial g_{o \nu}}{\partial x^{\mu}}+\frac{\partial g_{o \mu}}{\partial x^{\nu}}-\frac{\partial g_{\mu \nu}}{\partial x^{\sigma}}\right) \frac{d x^{\mu}}{d \lambda} \frac{d x^{\nu}}{d \lambda}=0 \tag{5} \end{equation*}(5)d2xαdλ2+gασ12(goνxμ+goμxνgμνxσ)dxμdλdxνdλ=0
which translates into the geodesic equation
(6) d 2 x α d λ 2 + Γ α μ ν d x μ d λ d x ν d λ = 0 (6) d 2 x α d λ 2 + Γ α μ ν d x μ d λ d x ν d λ = 0 {:(6)(d^(2)x^(alpha))/(dlambda^(2))+Gamma^(alpha)_(mu nu)(dx^(mu))/(d lambda)(dx^(nu))/(d lambda)=0:}\begin{equation*} \frac{d^{2} x^{\alpha}}{d \lambda^{2}}+\Gamma^{\alpha}{ }_{\mu \nu} \frac{d x^{\mu}}{d \lambda} \frac{d x^{\nu}}{d \lambda}=0 \tag{6} \end{equation*}(6)d2xαdλ2+Γαμνdxμdλdxνdλ=0
Thus, the new "dynamic" expression (2) is indeed extremal for geodesic curvesand, by contrast with proper length, (1), it is extremal when and only when the geodesic is affinely parametrized. [Its "Euler-Lagrange equations" (6) remain satisfied only under parameter changes λ new = a λ old + b λ new  = a λ old  + b lambda_("new ")=alambda_("old ")+b\lambda_{\text {new }}=a \lambda_{\text {old }}+bλnew =aλold +b, which keep the parameter affine; by contrast, the Euler-Lagrange equations (13.31) and (13.32) for the "principle of extremal length" (1) remain satisfied for any change of parameter whatsoever.]

Exercise 13.6. SPACELIKE GEODESICS HAVE EXTREMAL LENGTH

Show that any spacelike curve linking two events a a aaa and B B B\mathscr{B}B is a geodesic if and only if it extremizes the proper length
s = a B ( g μ ν d x μ d x ν ) 1 / 2 s = a B g μ ν d x μ d x ν 1 / 2 s=int_(a)^(B)(g_(mu nu)dx^(mu)dx^(nu))^(1//2)s=\int_{a}^{B}\left(g_{\mu \nu} d x^{\mu} d x^{\nu}\right)^{1 / 2}s=aB(gμνdxμdxν)1/2
[Hint: This is almost as easy as exercise 13.5 if one has already proved the analogous theorem for timelike geodesics.]

Exercise 13.7. METRIC TENSOR MEASURED BY LIGHT SIGNALS AND FREE PARTICLES [Kuchař]

(a) Instead of parametrizing a timelike geodesic by the proper time τ τ tau\tauτ, parametrize it by an arbitrary parameter μ μ mu\muμ,
τ = F ( μ ) τ = F ( μ ) tau=F(mu)\tau=F(\mu)τ=F(μ)
Write the geodesic equation in the μ μ mu\muμ-parametrization.
(b) Use now the coordinate time t t ttt as a parameter. Throw out a cloud of free particles with different "velocities" v i = d x i / d t v i = d x i / d t v^(i)=dx^(i)//dtv^{i}=d x^{i} / d tvi=dxi/dt and observe their "accelerations" a i = d 2 x i / d t 2 a i = d 2 x i / d t 2 a^(i)=d^(2)x^(i)//dt^(2)a^{i}=d^{2} x^{i} / d t^{2}ai=d2xi/dt2. Discuss what combinations of the components of the affine connection Γ i k λ Γ i k λ Gamma^(i)_(k lambda)\Gamma^{i}{ }_{k \lambda}Γikλ one can measure in this way. (Assume that no standard clocks measuring τ τ tau\tauτ are available!)
(c) Show that one can measure the conformal metric g ¯ u g ¯ u bar(g)_(u)\bar{g}_{u}g¯u, i.e., the ratios of the components of the metric tensor g i k g i k g_(ik)g_{i k}gik to a given component (say, g 00 g 00 g_(00)g_{00}g00 )
g ¯ u k = Λ g u k , Λ ( g 00 ) 1 g ¯ u k = Λ g u k , Λ g 00 1 bar(g)_(uk)=Lambdag_(uk),quad Lambda-=(-g_(00))^(-1)\bar{g}_{u k}=\Lambda g_{u k}, \quad \Lambda \equiv\left(-g_{00}\right)^{-1}g¯uk=Λguk,Λ(g00)1
using only the light signals moving along the null geodesics g ι k d x t d x κ = 0 g ι k d x t d x κ = 0 g_(iota k)dx^(t)dx^(kappa)=0g_{\iota k} d x^{t} d x^{\kappa}=0gιkdxtdxκ=0.
(d) Combine now the results of (b) and (c). Assume that Γ ι κ λ Γ ι κ λ Gamma^(iota)_(kappa lambda)\Gamma^{\iota}{ }_{\kappa \lambda}Γικλ is generated by the metric tensor by (13.22), (13.23), in the coordinate frame x t x t x^(t)x^{t}xt. Show that one can determine Λ Λ Lambda\LambdaΛ everywhere, if one prescribes it at one event (equivalent to fixing the unit of time).
Symmetries of Riemann in absence of metric
New symmetries imposed by metric

§13.5. METRIC-INDUCED PROPERTIES OF RIEMANN

In Newtonian spacetime, in the real, physical spacetime of Einstein-indeed, in any manifold with covariant derivative-the Riemann curvature tensor has these symmetries (exercise 11.6):
(13.40) R β γ δ α R β [ γ δ ] α (antisymmetry on last two indices) (13.41) R [ β γ δ ] α 0 (vanishing of completely antisymmetric part). (13.40) R β γ δ α R β [ γ δ ] α  (antisymmetry on last two indices)  (13.41) R [ β γ δ ] α 0  (vanishing of completely antisymmetric part).  {:[(13.40)R_(beta gamma delta)^(alpha)-=R_(beta[gamma delta])^(alpha)" (antisymmetry on last two indices) "],[(13.41)R_([beta gamma delta])^(alpha)-=0" (vanishing of completely antisymmetric part). "]:}\begin{align*} R_{\beta \gamma \delta}^{\alpha} & \equiv R_{\beta[\gamma \delta]}^{\alpha} & & \text { (antisymmetry on last two indices) } \tag{13.40}\\ R_{[\beta \gamma \delta]}^{\alpha} & \equiv 0 & & \text { (vanishing of completely antisymmetric part). } \tag{13.41} \end{align*}(13.40)RβγδαRβ[γδ]α (antisymmetry on last two indices) (13.41)R[βγδ]α0 (vanishing of completely antisymmetric part). 
In addition, it satisfies a differential identity (exercise 11.10):
(13.42) R β [ γ δ ; ϵ ] α 0 ("Bianchi identity") (13.42) R β [ γ δ ; ϵ ] α 0  ("Bianchi identity")  {:(13.42)R_(beta[gamma delta;epsilon])^(alpha)-=0quad" ("Bianchi identity") ":}\begin{equation*} R_{\beta[\gamma \delta ; \epsilon]}^{\alpha} \equiv 0 \quad \text { ("Bianchi identity") } \tag{13.42} \end{equation*}(13.42)Rβ[γδ;ϵ]α0 ("Bianchi identity") 
(see Chapter 15 for geometric significance).
When metric is brought onto the scene, whether in Einstein spacetime or elsewhere, it impresses on Riemann the additional symmetry (exercise 13.8)
(13.43) R α β γ δ R [ α β ] γ δ (antisymmetry on first two indices). (13.43) R α β γ δ R [ α β ] γ δ  (antisymmetry on first two indices).  {:(13.43)R_(alpha beta gamma delta)-=R_([alpha beta]gamma delta)" (antisymmetry on first two indices). ":}\begin{equation*} R_{\alpha \beta \gamma \delta} \equiv R_{[\alpha \beta] \gamma \delta} \text { (antisymmetry on first two indices). } \tag{13.43} \end{equation*}(13.43)RαβγδR[αβ]γδ (antisymmetry on first two indices). 
This, together with (13.40) and (13.41), forms a complete set of symmetries for Riemann; other symmetries that follow from these (exercise 13.10) are
(13.44) R α β γ δ = R γ δ α β (symmetry under pair exchange), (13.44) R α β γ δ = R γ δ α β  (symmetry under pair exchange),  {:(13.44)R_(alpha beta gamma delta)=R_(gamma delta alpha beta)quad" (symmetry under pair exchange), ":}\begin{equation*} R_{\alpha \beta \gamma \delta}=R_{\gamma \delta \alpha \beta} \quad \text { (symmetry under pair exchange), } \tag{13.44} \end{equation*}(13.44)Rαβγδ=Rγδαβ (symmetry under pair exchange), 
and
(13.45) R [ α β γ δ ] = 0 (vanishing of completely antisymmetric part). (13.45) R [ α β γ δ ] = 0  (vanishing of completely antisymmetric part).  {:(13.45)R_([alpha beta gamma delta])=0quad" (vanishing of completely antisymmetric part). ":}\begin{equation*} R_{[\alpha \beta \gamma \delta]}=0 \quad \text { (vanishing of completely antisymmetric part). } \tag{13.45} \end{equation*}(13.45)R[αβγδ]=0 (vanishing of completely antisymmetric part). 
These symmetries reduce the number of independent components of Riemann from 4 × 4 × 4 × 4 = 256 4 × 4 × 4 × 4 = 256 4xx4xx4xx4=2564 \times 4 \times 4 \times 4=2564×4×4×4=256 to 20 (exercise 13.9).
With metric present, one can construct a variety of new curvature tensors from Riemann. Some that will play important roles later are as follows.
(1) The double dual of Riemann, G G G-=^(**)\boldsymbol{G} \equiv{ }^{*}G Riemann* (analog of Maxwell -=\equiv
*Faraday), which has components
(13.46) ϵ α β γ δ 1 2 ϵ α β μ ν R μ ν ρ σ 1 2 ϵ ρ σ γ δ = 1 4 δ ρ σ γ δ α β μ ν R μ ν ρ σ (13.46) ϵ α β γ δ 1 2 ϵ α β μ ν R μ ν ρ σ 1 2 ϵ ρ σ γ δ = 1 4 δ ρ σ γ δ α β μ ν R μ ν ρ σ {:(13.46)epsilon^(alpha beta)_(gamma delta)-=(1)/(2)epsilon^(alpha beta mu nu)R_(mu nu)^(rho sigma)(1)/(2)epsilon_(rho sigma gamma delta)=-(1)/(4)delta_(rho sigma gamma delta)^(alpha beta mu nu)R_(mu nu)^(rho sigma):}\begin{equation*} \epsilon^{\alpha \beta}{ }_{\gamma \delta} \equiv \frac{1}{2} \epsilon^{\alpha \beta \mu \nu} R_{\mu \nu}^{\rho \sigma} \frac{1}{2} \epsilon_{\rho \sigma \gamma \delta}=-\frac{1}{4} \delta_{\rho \sigma \gamma \delta}^{\alpha \beta \mu \nu} R_{\mu \nu}{ }^{\rho \sigma} \tag{13.46} \end{equation*}(13.46)ϵαβγδ12ϵαβμνRμνρσ12ϵρσγδ=14δρσγδαβμνRμνρσ
(exercise 13.11).
(2) The Einstein curvature tensor, which is symmetric (exercise 13.11)
(13.47) G β δ G μ β μ δ ; G β δ = G δ β . (13.47) G β δ G μ β μ δ ; G β δ = G δ β . {:(13.47)G^(beta)_(delta)-=G^(mu beta)_(mu delta);quadG_(beta delta)=G_(delta beta).:}\begin{equation*} G^{\beta}{ }_{\delta} \equiv G^{\mu \beta}{ }_{\mu \delta} ; \quad G_{\beta \delta}=G_{\delta \beta} . \tag{13.47} \end{equation*}(13.47)GβδGμβμδ;Gβδ=Gδβ.
(3) The Ricci curvature tensor, which is symmetric, and the curvature scalar
(13.48) R β δ R μ β μ δ , R β δ = R δ β ; R R β β ; (13.48) R β δ R μ β μ δ , R β δ = R δ β ; R R β β ; {:(13.48)R^(beta)_(delta)-=R^(mu beta)_(mu delta)","quadR_(beta delta)=R_(delta beta);quad R-=R^(beta)_(beta);:}\begin{equation*} R^{\beta}{ }_{\delta} \equiv R^{\mu \beta}{ }_{\mu \delta}, \quad R_{\beta \delta}=R_{\delta \beta} ; \quad R \equiv R^{\beta}{ }_{\beta} ; \tag{13.48} \end{equation*}(13.48)RβδRμβμδ,Rβδ=Rδβ;RRββ;
which are related to the Einstein tensor by (exercise 13.12)
(13.49) R β δ = G β δ + 1 2 R δ β δ . (13.49) R β δ = G β δ + 1 2 R δ β δ . {:(13.49)R^(beta)_(delta)=G^(beta)_(delta)+(1)/(2)Rdelta^(beta)_(delta).:}\begin{equation*} R^{\beta}{ }_{\delta}=G^{\beta}{ }_{\delta}+\frac{1}{2} R \delta^{\beta}{ }_{\delta} . \tag{13.49} \end{equation*}(13.49)Rβδ=Gβδ+12Rδβδ.
(4) The Weyl conformal tensor (exercise 13.13)
(13.50) C α β γ δ = R α β γ δ 2 δ [ α [ γ R β ] δ ] + 1 3 δ [ α [ γ δ β ] δ ] R . (13.50) C α β γ δ = R α β γ δ 2 δ [ α [ γ R β ] δ ] + 1 3 δ [ α [ γ δ β ] δ ] R . {:(13.50)C^(alpha beta)_(gamma delta)=R^(alpha beta)_(gamma delta)-2delta^([alpha)_([gamma)R^(beta])_(delta])+(1)/(3)delta^([alpha)_([gamma)delta^(beta])_(delta])R.:}\begin{equation*} C^{\alpha \beta}{ }_{\gamma \delta}=R^{\alpha \beta}{ }_{\gamma \delta}-2 \delta^{[\alpha}{ }_{[\gamma} R^{\beta]}{ }_{\delta]}+\frac{1}{3} \delta^{[\alpha}{ }_{[\gamma} \delta^{\beta]}{ }_{\delta]} R . \tag{13.50} \end{equation*}(13.50)Cαβγδ=Rαβγδ2δ[α[γRβ]δ]+13δ[α[γδβ]δ]R.
The Bianchi identity (13.42) takes a particularly simple form when rewritten in terms of the double dual G G G\boldsymbol{G}G :
(13.51) G α β γ ; δ δ 0 ("Bianchi identity") (13.51) G α β γ ; δ δ 0  ("Bianchi identity")  {:(13.51)G_(alpha betagamma^(');delta)^(delta)-=0quad" ("Bianchi identity") ":}\begin{equation*} G_{\alpha \beta \gamma^{\prime} ; \delta}^{\delta} \equiv 0 \quad \text { ("Bianchi identity") } \tag{13.51} \end{equation*}(13.51)Gαβγ;δδ0 ("Bianchi identity") 
(exercise 13.11); and it has the obvious consequence
(13.52) G α β ; β 0 ("contracted Bianchi identity"). (13.52) G α β ; β 0  ("contracted Bianchi identity").  {:(13.52)G_(alpha)^(beta);beta-=0quad" ("contracted Bianchi identity"). ":}\begin{equation*} G_{\alpha}{ }^{\beta} ; \beta \equiv 0 \quad \text { ("contracted Bianchi identity"). } \tag{13.52} \end{equation*}(13.52)Gαβ;β0 ("contracted Bianchi identity"). 
Chapter 15 will be devoted to the deep geometric significance of these Bianchi identities.

EXERCISES

Exercise 13.8. RIEMANN ANTISYMMETRIC IN FIRST TWO INDICES

(a) Derive the antisymmetry condition (13.43). [Hint: Prove by abstract calculations that any vector fields s , u , v , w s , u , v , w s,u,v,w\boldsymbol{s}, \boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}s,u,v,w satisfy 0 = R ( u , v ) ( s w ) = s [ R ( u , v ) w ] + w [ R ( u , v ) s ] 0 = R ( u , v ) ( s w ) = s [ R ( u , v ) w ] + w [ R ( u , v ) s ] 0=R(u,v)(s*w)=s*[R(u,v)w]+w*[R(u,v)s]0=\mathscr{R}(\boldsymbol{u}, \boldsymbol{v})(\boldsymbol{s} \cdot \boldsymbol{w})=\boldsymbol{s} \cdot[\mathscr{R}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{w}]+\boldsymbol{w} \cdot[\mathscr{R}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{s}]0=R(u,v)(sw)=s[R(u,v)w]+w[R(u,v)s]. Then from this infer (13.43).]
(b) Explain in geometric terms the meaning of this antisymmetry.

Exercise 13.9. NUMBER OF INDEPENDENT COMPONENTS OF RIEMANN

(a) In the absence of metric, a complete set of symmetry conditions for Riemann is R α β γ 8 = R α β γ 8 = R^(alpha)_(beta gamma8)=R^{\alpha}{ }_{\beta \gamma 8}=Rαβγ8= R α β [ γ δ ] R α β [ γ δ ] R^(alpha)_(beta[gamma delta])R^{\alpha}{ }_{\beta[\gamma \delta]}Rαβ[γδ] and R α [ β γ δ ] = 0 R α [ β γ δ ] = 0 R^(alpha)_([beta gamma delta])=0R^{\alpha}{ }_{[\beta \gamma \delta]}=0Rα[βγδ]=0. Show that in four-dimensional spacetime these reduce the number of independent components from 4 × 4 × 4 × 4 = 256 4 × 4 × 4 × 4 = 256 4xx4xx4xx4=2564 \times 4 \times 4 \times 4=2564×4×4×4=256 to 4 × 4 × 6 4 × 4 = 96 4 × 4 × 6 4 × 4 = 96 4xx4xx6-4xx4=96-4 \times 4 \times 6-4 \times 4=96-4×4×64×4=96 16 = 80 16 = 80 16=8016=8016=80.
(b) Show that in a manifold of n n nnn dimensions without metric, the number of independent components is
(13.53) n 3 ( n 1 ) 2 n 2 ( n 1 ) ( n 2 ) 6 = n 2 ( n 2 1 ) 3 (13.53) n 3 ( n 1 ) 2 n 2 ( n 1 ) ( n 2 ) 6 = n 2 n 2 1 3 {:(13.53)(n^(3)(n-1))/(2)-(n^(2)(n-1)(n-2))/(6)=(n^(2)(n^(2)-1))/(3):}\begin{equation*} \frac{n^{3}(n-1)}{2}-\frac{n^{2}(n-1)(n-2)}{6}=\frac{n^{2}\left(n^{2}-1\right)}{3} \tag{13.53} \end{equation*}(13.53)n3(n1)2n2(n1)(n2)6=n2(n21)3
(c) In the presence of metric, a complete set of symmetries is R α β γ δ = R [ α β ] [ γ δ ] R α β γ δ = R [ α β ] [ γ δ ] R_(alpha beta gamma delta)=R_([alpha beta][gamma delta])R_{\alpha \beta \gamma \delta}=R_{[\alpha \beta][\gamma \delta]}Rαβγδ=R[αβ][γδ], and R α [ β γ δ ] = 0 R α [ β γ δ ] = 0 R_(alpha[beta gamma delta])=0R_{\alpha[\beta \gamma \delta]}=0Rα[βγδ]=0. Show that in four-dimensional spacetime, these reduce the number of independent components to 6 × 6 4 × 4 = 36 16 = 20 6 × 6 4 × 4 = 36 16 = 20 6xx6-4xx4=36-16=206 \times 6-4 \times 4=36-16=206×64×4=3616=20.
(d) Show that in a manifold of n n nnn dimensions with metric, the number of independent components is
(13.54) [ n ( n 1 ) 2 ] 2 n 2 ( n 1 ) ( n 2 ) 6 = n 2 ( n 2 1 ) 12 (13.54) n ( n 1 ) 2 2 n 2 ( n 1 ) ( n 2 ) 6 = n 2 n 2 1 12 {:(13.54)[(n(n-1))/(2)]^(2)-(n^(2)(n-1)(n-2))/(6)=(n^(2)(n^(2)-1))/(12):}\begin{equation*} \left[\frac{n(n-1)}{2}\right]^{2}-\frac{n^{2}(n-1)(n-2)}{6}=\frac{n^{2}\left(n^{2}-1\right)}{12} \tag{13.54} \end{equation*}(13.54)[n(n1)2]2n2(n1)(n2)6=n2(n21)12

Exercise 13.10. RIEMANN SYMMETRIC IN EXCHANGE OF PAIRS; COMPLETELY ANTISYMMETRIC PART VANISHES

From the complete set of symmetries in the presence of a metric, R α β γ δ = R [ α β ] [ γ δ ] R α β γ δ = R [ α β ] [ γ δ ] R_(alpha beta gamma delta)=R_([alpha beta][gamma delta])R_{\alpha \beta \gamma \delta}=R_{[\alpha \beta][\gamma \delta]}Rαβγδ=R[αβ][γδ] and R α [ β γ δ ] = 0 R α [ β γ δ ] = 0 R_(alpha[beta gamma delta])=0R_{\alpha[\beta \gamma \delta]}=0Rα[βγδ]=0, derive: (a) symmetry under pair exchange, R α β γ δ = R γ δ α β R α β γ δ = R γ δ α β R_(alpha beta gamma delta)=R_(gamma delta alpha beta)R_{\alpha \beta \gamma \delta}=R_{\gamma \delta \alpha \beta}Rαβγδ=Rγδαβ, and (b) vanishing of completely antisymmetric part, R [ α β γ δ ] = 0 R [ α β γ δ ] = 0 R_([alpha beta gamma delta])=0R_{[\alpha \beta \gamma \delta]}=0R[αβγδ]=0. Then (c) show that the following form a complete set of symmetries:
(13.55) R α β γ δ = R [ α β ] [ γ δ ] = R γ δ α β , R [ α β γ δ ] = 0 (13.55) R α β γ δ = R [ α β ] [ γ δ ] = R γ δ α β , R [ α β γ δ ] = 0 {:(13.55)R_(alpha beta gamma delta)=R_([alpha beta][gamma delta])=R_(gamma delta alpha beta)","quadR_([alpha beta gamma delta])=0:}\begin{equation*} R_{\alpha \beta \gamma \delta}=R_{[\alpha \beta][\gamma \delta]}=R_{\gamma \delta \alpha \beta}, \quad R_{[\alpha \beta \gamma \delta]}=0 \tag{13.55} \end{equation*}(13.55)Rαβγδ=R[αβ][γδ]=Rγδαβ,R[αβγδ]=0

Exercise 13.11. DOUBLE DUAL OF RIEMANN; EINSTEIN

(a) Show that G G G-=^(**)G \equiv{ }^{*}G Riemann* contains precisely the same amount of information as Riemann, and satisfies precisely the same set of symmetries [(13.40), (13.41), (13.43) to (13.45)].
(b) From the symmetries of G G G\boldsymbol{G}G, show that Einstein [defined in (13.47)] is symmetric ( G [ β δ ] = 0 G [ β δ ] = 0 G_([beta delta])=0G_{[\beta \delta]}=0G[βδ]=0 ).
(c) Show that the Bianchi identities (13.42), when written in terms of G G G\boldsymbol{G}G, take the form (13.51) ("vanishing divergence," G = 0 G = 0 grad*G=0\boldsymbol{\nabla} \cdot \boldsymbol{G}=0G=0 ).
(d) By contracting the Bianchi identities G = 0 G = 0 grad*G=0\boldsymbol{\nabla} \cdot \boldsymbol{G}=0G=0, show that G G G-=\boldsymbol{G} \equivG Einstein has vanishing divergence [equation (13.52)].
(a) From the symmetries of Riemann, show that Ricci is symmetric ( R [ β δ ] = 0 ) R [ β δ ] = 0 (R_([beta delta])=0)\left(R_{[\beta \delta]}=0\right)(R[βδ]=0).
(b) Show that Ricci is related to Einstein by equation (13.49).

Exercise 13.13. THE WEYL CONFORMAL TENSOR

(a) Show that the Weyl conformal tensor (13.50) possesses the same symmetries [(13.40), (13.41), (13.43) to (13.45)] as the Riemann tensor.
(b) Show that the Weyl tensor is completely "trace-free"; i.e., that
(13.56) contraction of C α β γ δ on any pair of slots vanishes. (13.56)  contraction of  C α β γ δ  on any pair of slots vanishes.  {:(13.56)" contraction of "C_(alpha beta gamma delta)" on any pair of slots vanishes. ":}\begin{equation*} \text { contraction of } C_{\alpha \beta \gamma \delta} \text { on any pair of slots vanishes. } \tag{13.56} \end{equation*}(13.56) contraction of Cαβγδ on any pair of slots vanishes. 
Thus, C α β γ δ C α β γ δ C_(alpha beta gamma delta)C_{\alpha \beta \gamma \delta}Cαβγδ can be regarded as the trace-free part of Riemann, and R α β R α β R_(alpha beta)R_{\alpha \beta}Rαβ can be regarded as the trace of Riemann. Riemann is determined entirely by its trace-free part C α β γ δ C α β γ δ C_(alpha beta gamma delta)C_{\alpha \beta \gamma \delta}Cαβγδ and its trace R α β R α β R_(alpha beta)R_{\alpha \beta}Rαβ [see equation (13.50), and recall R = R α α R = R α α R=R^(alpha)_(alpha)R=R^{\alpha}{ }_{\alpha}R=Rαα ].
(c) Show that in spacetime the Weyl tensor has 10 independent components.
(d) Show that in an n n nnn-dimensional manifold the number of independent components of Weyl [defined by a modification of (13.50) that maintains (13.56)] is
(13.57) n 2 ( n 2 1 ) 12 n ( n + 1 ) 2 for n 3 , 0 for n 3 (13.57) n 2 n 2 1 12 n ( n + 1 ) 2  for  n 3 , 0  for  n 3 {:(13.57)(n^(2)(n^(2)-1))/(12)-(n(n+1))/(2)" for "n >= 3","quad0" for "n <= 3:}\begin{equation*} \frac{n^{2}\left(n^{2}-1\right)}{12}-\frac{n(n+1)}{2} \text { for } n \geq 3, \quad 0 \text { for } n \leq 3 \tag{13.57} \end{equation*}(13.57)n2(n21)12n(n+1)2 for n3,0 for n3
Thus, in manifolds of 1,2 , or 3 dimensions, the Weyl tensor is identically zero, and the Ricci tensor completely determines the Riemann tensor.

§13.6. THE PROPER REFERENCE FRAME OF AN ACCELERATED OBSERVER

A physicist performing an experiment in a jet airplane (e.g., an infrared astronomy experiment) may use several different coordinate systems at once. But a coordinate system of special utility is one at rest relative to all the apparatus bolted into the floor and walls of the airplane cabin. This "proper reference frame" has a rectangular " x ^ , y ^ , z ^ x ^ , y ^ , z ^ hat(x), hat(y), hat(z)\hat{x}, \hat{y}, \hat{z}x^,y^,z^ " grid attached to the walls of the cabin, and one or more clocks at rest in the grid. That this proper reference frame is accelerated relative to the local Lorentz frames, the physicist knows from his own failure to float freely in the cabin, or, with greater precision, from accelerometer measurements. That his proper reference frame is rotating relative to local Lorentz frames he knows from the Coriolis forces he feels, or, with greater precision, from the rotation of inertial-guidance gyroscopes relative to the cabin walls.
Exercise 6.8 gave a mathematical treatment of such an accelerated, rotating, but locally orthonormal reference frame in flat spacetime. This section does the same in curved spacetime. In the immediate vicinity of the spatial grid's origin x j ^ = 0 x j ^ = 0 x^( hat(j))=0x^{\hat{j}}=0xj^=0 (region of spatial extent so small that curvature effects are negligible), no aspect of the coordinate system can possibly reveal whether spacetime is curved or flat. Hence, all the details of exercise 6.8 must remain valid in curved spacetime. Nevertheless, it is instructive to rediscuss those details, and some new ones, using the powerful mathematics of the last few chapters.
Begin by making more precise the coordinate grid to be used. The following is perhaps the most natural way to set up the grid.
(1) Let τ τ tau\tauτ be proper time as measured by the accelerated observer's clock (clock at center of airplane cabin in above example). Let P = P 0 ( τ ) P = P 0 ( τ ) P=P_(0)(tau)\mathscr{P}=\mathscr{P}_{0}(\tau)P=P0(τ) be the observer's world line, as shown in Figure 13.4,a.
Proper reference frame described physically
Six-step construction of coordinate grid for proper frame
Figure 13:4.
The proper reference frame of an accelerated observer. Diagram (a) shows the observer's orthonormal tetrad { e a } e a {e_(a)}\left\{\boldsymbol{e}_{a}\right\}{ea} being transported along his world line P 0 ( τ ) P 0 ( τ ) P_(0)(tau)\mathscr{P}_{0}(\tau)P0(τ) [transport law (13.60)]. Diagram (b) shows geodesics bristling out perpendicularly from an arbitrary event P 0 ( 4 ) P 0 ( 4 ) P_(0)(4)\mathscr{\mathscr { P }}_{0}(4)P0(4) on the observer's world line. Each geodesic is specified uniquely by (1) the proper time τ τ tau\tauτ at which it originates, and (2) the direction (unit tangent vector n = d / d s = n j e j n = d / d s = n j e j n=d//ds=n^(j)e_(j)\boldsymbol{n}=d / d s=n^{j} \boldsymbol{e}_{j}n=d/ds=njej along which it emanates). A given event on the geodesic is specified by τ , n τ , n tau,n\tau, \boldsymbol{n}τ,n, and proper distance s s sss from the geodesic's emanation point; hence the notation
P = G [ τ , n , s ] P = G [ τ , n , s ] P=G[tau,n,s]\mathscr{P}=\mathscr{G}[\tau, \boldsymbol{n}, s]P=G[τ,n,s]
for the given event. The observer's proper reference frame attributes to this given event the coordinates
x 0 ^ ( G [ τ , n , s ] ) = τ , x j ( G [ τ , n , s ] ) = s n j ^ . x 0 ^ ( G [ τ , n , s ] ) = τ , x j ( G [ τ , n , s ] ) = s n j ^ . {:[x^( hat(0)(G[tau,n,s])=tau,)],[x^(j)(G[tau","n","s])=sn^( hat(j)).]:}\begin{aligned} & x^{\hat{0}(\mathscr{G}[\tau, \boldsymbol{n}, s])=\tau,} \\ & x^{j}(\mathscr{G}[\tau, \boldsymbol{n}, s])=s n^{\hat{j}} . \end{aligned}x0^(G[τ,n,s])=τ,xj(G[τ,n,s])=snj^.
(2) The observer carries with himself an orthonormal tetrad { e α ^ } e α ^ {e_( hat(alpha))}\left\{\boldsymbol{e}_{\hat{\alpha}}\right\}{eα^} (Figure 13.4,a), with
(13.58) e 0 ^ = u = d P 0 / d τ = ( 4 -velocity of observer ) (13.58) e 0 ^ = u = d P 0 / d τ = ( 4 -velocity of observer  ) {:(13.58)e_( hat(0))=u=dP_(0)//d tau=(4"-velocity of observer "):}\begin{equation*} \boldsymbol{e}_{\hat{0}}=\boldsymbol{u}=d \mathscr{P}_{0} / d \tau=(4 \text {-velocity of observer }) \tag{13.58} \end{equation*}(13.58)e0^=u=dP0/dτ=(4-velocity of observer )
( e 0 ^ e 0 ^ e_( hat(0))\boldsymbol{e}_{\hat{0}}e0^ points along observer's "time direction"), and with
(13.59) e α ^ e β ^ = η α β (13.59) e α ^ e β ^ = η α β {:(13.59)e_( hat(alpha))*e_( hat(beta))=eta_(alpha beta):}\begin{equation*} \boldsymbol{e}_{\hat{\alpha}} \cdot \boldsymbol{e}_{\hat{\beta}}=\eta_{\alpha \beta} \tag{13.59} \end{equation*}(13.59)eα^eβ^=ηαβ
(orthonormality).
(3) The tetrad changes from point to point along the observer's world line, relative to parallel transport:
(13.60) u e α ^ = Ω e α ^ , (13.61) Ω μ ν = a μ u ν u μ a ν + u α ω β ϵ α β μ ν = "generator of infinitesimal Lorentz transformation." (13.60) u e α ^ = Ω e α ^ , (13.61) Ω μ ν = a μ u ν u μ a ν + u α ω β ϵ α β μ ν =  "generator of infinitesimal Lorentz transformation."  {:[(13.60)grad_(u)e_( hat(alpha))=-Omega*e_( hat(alpha))","],[(13.61)Omega^(mu nu)=a^(mu)u^(nu)-u^(mu)a^(nu)+u_(alpha)omega_(beta)epsilon^(alpha beta mu nu)],[=" "generator of infinitesimal Lorentz transformation." "]:}\begin{align*} & \boldsymbol{\nabla}_{u} \boldsymbol{e}_{\hat{\alpha}}=-\boldsymbol{\Omega} \cdot \boldsymbol{e}_{\hat{\alpha}}, \tag{13.60}\\ & \Omega^{\mu \nu}=a^{\mu} u^{\nu}-u^{\mu} a^{\nu}+u_{\alpha} \omega_{\beta} \epsilon^{\alpha \beta \mu \nu} \tag{13.61}\\ &=\text { "generator of infinitesimal Lorentz transformation." } \end{align*}(13.60)ueα^=Ωeα^,(13.61)Ωμν=aμuνuμaν+uαωβϵαβμν= "generator of infinitesimal Lorentz transformation." 
This transport law has the same form in curved spacetime as in flat ( $ 6.5 $ 6.5 $6.5\$ 6.5$6.5 and exercise 6.8 ) because curvature can only be felt over finite distances, not over the infinitesimal distance involved in the "first time-rate of change of a vector" (equivalence principle). As in exercise 6.8,
a = u u = ( 4 acceleration of observer ) , (13.62) ω = ( angular velocity of rotation of spatial basis vectors e j relative to Fermi- Walker-transported vectors, i.e., relative to inertial-guidance gyroscopes ) , u a = u ω = 0 . a = u u = ( 4  acceleration of observer  ) , (13.62) ω =  angular velocity of rotation of spatial   basis vectors  e j  relative to Fermi-   Walker-transported vectors, i.e.,   relative to inertial-guidance gyroscopes  , u a = u ω = 0 . {:[a=grad_(u)u=(4-" acceleration of observer ")","],[(13.62)omega=([" angular velocity of rotation of spatial "],[" basis vectors "e_(j)" relative to Fermi- "],[" Walker-transported vectors, i.e., "],[" relative to inertial-guidance gyroscopes "])","],[u*a=u*omega=0.]:}\begin{align*} \boldsymbol{a} & =\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=(4-\text { acceleration of observer }), \\ \boldsymbol{\omega} & =\left(\begin{array}{l} \text { angular velocity of rotation of spatial } \\ \text { basis vectors } \boldsymbol{e}_{j} \text { relative to Fermi- } \\ \text { Walker-transported vectors, i.e., } \\ \text { relative to inertial-guidance gyroscopes } \end{array}\right), \tag{13.62}\\ \boldsymbol{u} \cdot \boldsymbol{a} & =\boldsymbol{u} \cdot \boldsymbol{\omega}=0 . \end{align*}a=uu=(4 acceleration of observer ),(13.62)ω=( angular velocity of rotation of spatial  basis vectors ej relative to Fermi-  Walker-transported vectors, i.e.,  relative to inertial-guidance gyroscopes ),ua=uω=0.
If ω ω omega\boldsymbol{\omega}ω were zero, the observer would be Fermi-Walker-transporting his tetrad (gyroscope-type transport). If both a a a\boldsymbol{a}a and ω ω omega\boldsymbol{\omega}ω were zero, he would be freely falling (geodesic motion) and would be parallel-transporting his tetrad, u e α ^ = 0 u e α ^ = 0 grad_(u)e_( hat(alpha))=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{e}_{\hat{\alpha}}=0ueα^=0.
(4) The observer constructs his proper reference frame (local coordinate system) in a manner analogous to the Riemann-normal construction of § 11.6 § 11.6 §11.6\S 11.6§11.6. From each event P 0 ( τ ) P 0 ( τ ) P_(0)(tau)\mathscr{P}_{0}(\tau)P0(τ) on his world line, he sends out purely spatial geodesics (geodesics orthogonal to u = d P 0 / d τ ) u = d P 0 / d τ {:u=dP_(0)//d tau)\left.\boldsymbol{u}=d \mathscr{P}{ }_{0} / d \tau\right)u=dP0/dτ), with affine parameter equal to proper length,

(See Figure 13.4,b.) The tangent vector has unit length, because the chosen affine parameter is proper length:
(13.64) n = ( G / s ) s = 0 ; n μ = ( d x μ / d s ) along geodesic, n n = g μ v ( d x μ d s ) ( d x ν d s ) = d s 2 d s 2 = 1 (13.64) n = ( G / s ) s = 0 ; n μ = d x μ / d s  along geodesic,  n n = g μ v d x μ d s d x ν d s = d s 2 d s 2 = 1 {:[(13.64)n=(delG//del s)_(s)=0;quadn^(mu)=(dx^(mu)//ds)" along geodesic, "],[n*n=g_(mu v)((dx^(mu))/(ds))((dx^(nu))/(ds))=(ds^(2))/(ds^(2))=1]:}\begin{align*} \boldsymbol{n} & =(\partial \mathscr{G} / \partial s)_{s}=0 ; \quad n^{\mu}=\left(d x^{\mu} / d s\right) \text { along geodesic, } \tag{13.64}\\ \boldsymbol{n} \cdot \boldsymbol{n} & =g_{\mu v}\left(\frac{d x^{\mu}}{d s}\right)\left(\frac{d x^{\nu}}{d s}\right)=\frac{d s^{2}}{d s^{2}}=1 \end{align*}(13.64)n=(G/s)s=0;nμ=(dxμ/ds) along geodesic, nn=gμv(dxμds)(dxνds)=ds2ds2=1
(5) Each event near the observer's world line is intersected by precisely one of the geodesics G [ τ , n , s ] G [ τ , n , s ] G[tau,n,s]\mathcal{G}[\tau, \boldsymbol{n}, s]G[τ,n,s]. [Far away, this is not true; the geodesics may cross, either because of the observer's acceleration, as in Figure 6.3, or because of the curvature of spacetime ("geodesic deviation").]
(6) Pick an event P P P\mathscr{P}P near the observer's world line. The geodesic through it originated on the observer's world line at a specific time τ τ tau\tauτ, had original direction n = n β ^ e j ^ n = n β ^ e j ^ n=n^( hat(beta))e_( hat(j))\boldsymbol{n}=n^{\hat{\beta}} \boldsymbol{e}_{\hat{j}}n=nβ^ej^, and needed to extend a distance s s sss before reaching P P P\mathscr{P}P. Hence, the four numbers
(13.65) ( x 0 ^ , x 1 ^ , x 2 ^ , x 3 ^ ) ( τ , s n 1 ^ , s n 2 ^ , s n 3 ^ ) (13.65) x 0 ^ , x 1 ^ , x 2 ^ , x 3 ^ τ , s n 1 ^ , s n 2 ^ , s n 3 ^ {:(13.65)(x^( hat(0)),x^( hat(1)),x^( hat(2)),x^( hat(3)))-=(tau,sn^( hat(1)),sn^( hat(2)),sn^( hat(3))):}\begin{equation*} \left(x^{\hat{0}}, x^{\hat{1}}, x^{\hat{2}}, x^{\hat{3}}\right) \equiv\left(\tau, s n^{\hat{1}}, s n^{\hat{2}}, s n^{\hat{3}}\right) \tag{13.65} \end{equation*}(13.65)(x0^,x1^,x2^,x3^)(τ,sn1^,sn2^,sn3^)
are a natural way of identifying the event P P P\mathscr{P}P. These are the coordinates of P P P\mathscr{P}P in the observer's proper reference frame.
(7) Restated more abstractly,
(13.65') x 0 ^ ( G [ τ , n , s ] ) = τ , x j ^ ( G [ τ , n , s ] ) = s n j ^ = s n j ^ = s n e j ^ . (13.65') x 0 ^ ( G [ τ , n , s ] ) = τ , x j ^ ( G [ τ , n , s ] ) = s n j ^ = s n j ^ = s n e j ^ . {:[(13.65')x^( hat(0))(G[tau","n","s])=tau","],[x^( hat(j))(G[tau","n","s])=sn^( hat(j))=sn_( hat(j))=sn*e_( hat(j)).]:}\begin{align*} x^{\hat{0}}(\mathscr{G}[\tau, \boldsymbol{n}, s]) & =\tau, \tag{13.65'}\\ x^{\hat{j}}(\mathscr{G}[\tau, \boldsymbol{n}, s]) & =s n^{\hat{j}}=s n_{\hat{j}}=s \boldsymbol{n} \cdot \boldsymbol{e}_{\hat{j}} . \end{align*}(13.65')x0^(G[τ,n,s])=τ,xj^(G[τ,n,s])=snj^=snj^=snej^.
In flat spacetime this construction process and the resulting coordinates x α ^ ( P ) x α ^ ( P ) x^( hat(alpha))(P)x^{\hat{\alpha}}(\mathscr{P})xα^(P) are identical to the process and resulting coordinates ξ α ( P ) ξ α ( P ) xi^(alpha^('))(P)\xi^{\alpha^{\prime}}(\mathscr{P})ξα(P) of exercise 6.8.
For use in calculations one wants not only the coordinate system, but also its metric coefficients and connection coefficients. Fortunately, g α ^ β ^ g α ^ β ^ g_( hat(alpha) hat(beta))g_{\hat{\alpha} \hat{\beta}}gα^β^ and Γ α ^ β ^ γ ^ Γ α ^ β ^ γ ^ Gamma^( hat(alpha))_( hat(beta) hat(gamma))\Gamma^{\hat{\alpha}}{ }_{\hat{\beta} \hat{\gamma}}Γα^β^γ^ are needed only along the observer's world line, where they are especially simple. Only a foolish observer would try to use his own proper reference frame far from his world line, where its grid ceases to be orthonormal and its geodesic grid lines may even cross! (See §6.3.)
All along the observer's world line P 0 ( τ ) P 0 ( τ ) P_(0)(tau)\mathscr{\mathscr { P }}_{0}(\tau)P0(τ), the basis vectors of his coordinate grid are identical (by construction) to his orthonormal tetrad
(13.66) / x α ^ = e α ^ (13.66) / x α ^ = e α ^ {:(13.66)del//delx^( hat(alpha))=e_( hat(alpha)):}\begin{equation*} \partial / \partial x^{\hat{\alpha}}=\boldsymbol{e}_{\hat{\alpha}} \tag{13.66} \end{equation*}(13.66)/xα^=eα^
and therefore its metric coefficients are
(13.67) g α ^ β ^ = e α ^ e β ^ = η α β all along P 0 ( τ ) . (13.67) g α ^ β ^ = e α ^ e β ^ = η α β  all along  P 0 ( τ ) . {:(13.67)g_( hat(alpha) hat(beta))=e_( hat(alpha))*e_( hat(beta))=eta_(alpha beta)" all along "P_(0)(tau).:}\begin{equation*} g_{\hat{\alpha} \hat{\beta}}=\boldsymbol{e}_{\hat{\alpha}} \cdot \boldsymbol{e}_{\hat{\beta}}=\eta_{\alpha \beta} \text { all along } \mathscr{P}_{0}(\tau) . \tag{13.67} \end{equation*}(13.67)gα^β^=eα^eβ^=ηαβ all along P0(τ).
Some of the connection coefficients are determined by the transport law (13.60) for the observer's orthonormal tetrad:
u e α ^ = 0 ^ e α ^ = e β Γ β ^ α ^ 0 ^ ^ = Ω e α ^ = e β ^ Ω β ^ α ^ . u e α ^ = 0 ^ e α ^ = e β Γ β ^ α ^ 0 ^ ^ = Ω e α ^ = e β ^ Ω β ^ α ^ . {:[grad_(u)e_( hat(alpha))=grad_( hat(0))e_( hat(alpha))=e_(beta)Gamma^( hat(beta)) hat(alpha) hat(hat(0))],[=-Omega*e_( hat(alpha))=-e_( hat(beta))Omega^( hat(beta))_( hat(alpha)).]:}\begin{aligned} \boldsymbol{\nabla}_{u} \boldsymbol{e}_{\hat{\alpha}} & =\boldsymbol{\nabla}_{\hat{0}} \boldsymbol{e}_{\hat{\alpha}}=\boldsymbol{e}_{\beta} \Gamma^{\hat{\beta}} \hat{\alpha} \hat{\hat{0}} \\ & =-\boldsymbol{\Omega} \cdot \boldsymbol{e}_{\hat{\alpha}}=-\boldsymbol{e}_{\hat{\beta}} \Omega^{\hat{\beta}}{ }_{\hat{\alpha}} . \end{aligned}ueα^=0^eα^=eβΓβ^α^0^^=Ωeα^=eβ^Ωβ^α^.
Thus
(13.68) Γ β ^ α ^ 0 ^ = Ω β ^ α ^ α ^ all along P 0 ( τ ) . (13.68) Γ β ^ α ^ 0 ^ = Ω β ^ α ^ α ^  all along  P 0 ( τ ) . {:(13.68)Gamma^( hat(beta))_( hat(alpha) hat(0))=-Omega^( hat(beta)) hat(alpha)_( hat(alpha))" all along "P_(0)(tau).:}\begin{equation*} \Gamma^{\hat{\beta}}{ }_{\hat{\alpha} \hat{0}}=-\Omega^{\hat{\beta}} \hat{\alpha}_{\hat{\alpha}} \text { all along } \mathscr{P}_{0}(\tau) . \tag{13.68} \end{equation*}(13.68)Γβ^α^0^=Ωβ^α^α^ all along P0(τ).
Since Ω Ω Omega\OmegaΩ has the form (13.61) and the observer's 4 -velocity and 4 -acceleration have components u 0 ^ = 1 , u j ^ = 0 , a o ^ = 0 u 0 ^ = 1 , u j ^ = 0 , a o ^ = 0 u_( hat(0))=-1,u_( hat(j))=0,a_( hat(o))=0u_{\hat{0}}=-1, u_{\hat{j}}=0, a_{\hat{o}}=0u0^=1,uj^=0,ao^=0 in the observer's own proper frame, these connection coefficients are
The remaining connection coefficients can be read from the geodesic equation for the geodesics G [ τ , n , s ] G [ τ , n , s ] G[tau,n,s]\mathscr{G}[\tau, \boldsymbol{n}, s]G[τ,n,s] that emanate from the observer's world line. According to equation (13.65), the coordinate representation of each such geodesic is
x 0 ^ ( s ) = τ = constant , x ȷ ^ ( s ) = n ı ^ s ; x 0 ^ ( s ) = τ =  constant  , x ȷ ^ ( s ) = n ı ^ s ; x^( hat(0))(s)=tau=" constant ",quadx^( hat(ȷ))(s)=n^( hat(ı))s;x^{\hat{0}}(s)=\tau=\text { constant }, \quad x^{\hat{\jmath}}(s)=n^{\hat{\imath}} s ;x0^(s)=τ= constant ,xȷ^(s)=nı^s;
hence, d 2 x α ^ / d s 2 = 0 d 2 x α ^ / d s 2 = 0 d^(2)x^( hat(alpha))//ds^(2)=0d^{2} x^{\hat{\alpha}} / d s^{2}=0d2xα^/ds2=0 all along the geodesic, and the geodesic equation reads
0 = d 2 x α ^ d s 2 + Γ α ^ β β ^ γ ^ d x β ^ d s d x γ ^ d s = Γ α ^ j ^ k ^ n β ^ n k ^ . 0 = d 2 x α ^ d s 2 + Γ α ^ β β ^ γ ^ d x β ^ d s d x γ ^ d s = Γ α ^ j ^ k ^ n β ^ n k ^ . 0=(d^(2)x^( hat(alpha)))/(ds^(2))+Gamma^( hat(alpha))_(beta hat(beta) hat(gamma))(dx^( hat(beta)))/(ds)(dx^( hat(gamma)))/(ds)=Gamma^( hat(alpha))_( hat(j) hat(k))n^( hat(beta))n^( hat(k)).0=\frac{d^{2} x^{\hat{\alpha}}}{d s^{2}}+\Gamma^{\hat{\alpha}}{ }_{\beta \hat{\beta} \hat{\gamma}} \frac{d x^{\hat{\beta}}}{d s} \frac{d x^{\hat{\gamma}}}{d s}=\Gamma^{\hat{\alpha}}{ }_{\hat{j} \hat{k}} n^{\hat{\beta}} n^{\hat{k}} .0=d2xα^ds2+Γα^ββ^γ^dxβ^dsdxγ^ds=Γα^j^k^nβ^nk^.
This equation is satisfied on the observer's world line for all spatial geodesics (all n 3 ^ n 3 ^ n^( hat(3))n^{\hat{3}}n3^ ) if and only if
(13.69b) Γ α ^ j k ^ = Γ α ^ j ^ k = 0 all along P 0 ( τ ) . (13.69b) Γ α ^ j k ^ = Γ α ^ j ^ k = 0  all along  P 0 ( τ ) . {:(13.69b)Gamma^( hat(alpha)_(j hat(k)))=Gamma_( hat(alpha) hat(j)k)=0" all along "P_(0)(tau).:}\begin{equation*} \Gamma^{\hat{\alpha}_{j \hat{k}}}=\Gamma_{\hat{\alpha} \hat{j} k}=0 \text { all along } \mathscr{P}_{0}(\tau) . \tag{13.69b} \end{equation*}(13.69b)Γα^jk^=Γα^j^k=0 all along P0(τ).
The values (13.69) of the connection coefficients determine uniquely the partial derivatives of the metric coefficients [see equation (13.19')]:
(13.70) g α ^ β , 0 ^ = 0 , g j k , l ^ , = 0 , g 0 ^ 0 ^ , j ^ = 2 a j , g 0 ^ j ^ , k ^ = ϵ 0 ^ ȷ ^ ω } } all along P 0 ( τ ) ; (13.70) g α ^ β , 0 ^ = 0 , g j k , l ^ , = 0 , g 0 ^ 0 ^ , j ^ = 2 a j , g 0 ^ j ^ , k ^ = ϵ 0 ^ ȷ ^ ω  all along  P 0 ( τ ) ; {:(13.70){:[g_( hat(alpha)beta, hat(0)),=0",",,g_( hat(jk,l),)=0","],[g_( hat(0) hat(0), hat(j)),=-2a_(j)",",,{:g_( hat(0) hat(j), hat(k))=-epsilon_( hat(0) hat(ȷ)ℓℓ)omega^(ℓ)}]}" all along "P_(0)(tau);:}\left.\begin{array}{rlrl} g_{\hat{\alpha} \beta, \hat{0}} & =0, & & g_{\hat{j k, l},}=0, \tag{13.70}\\ g_{\hat{0} \hat{0}, \hat{j}} & =-2 a_{j}, & & \left.g_{\hat{0} \hat{j}, \hat{k}}=-\epsilon_{\hat{0} \hat{\jmath} \ell \ell} \omega^{\ell}\right\} \end{array}\right\} \text { all along } \mathscr{P}_{0}(\tau) ;(13.70)gα^β,0^=0,gjk,l^,=0,g0^0^,j^=2aj,g0^j^,k^=ϵ0^ȷ^ω}} all along P0(τ);
and these, plus the orthonormality condition g α ^ β ^ [ P 0 ( τ ) ] = η α β g α ^ β ^ P 0 ( τ ) = η α β g_( hat(alpha) hat(beta))[P_(0)(tau)]=eta_(alpha beta)g_{\hat{\alpha} \hat{\beta}}\left[\mathscr{P}_{0}(\tau)\right]=\eta_{\alpha \beta}gα^β^[P0(τ)]=ηαβ, imply that the line element near the observer's world line is
d s 2 = ( 1 + 2 a j ^ x ȷ ^ ) d x ^ 2 2 ( ϵ j k ^ x k ^ ω ^ ) d x 0 ^ d x 3 ^ (13.71) + δ j j k ^ d x 3 ^ d x k ^ + O ( | x 3 ^ | 2 ) d x α ^ d x β ^ . d s 2 = 1 + 2 a j ^ x ȷ ^ d x ^ 2 2 ϵ j k ^ x k ^ ω ^ d x 0 ^ d x 3 ^ (13.71) + δ j j k ^ d x 3 ^ d x k ^ + O x 3 ^ 2 d x α ^ d x β ^ . {:[ds^(2)=-(1+2a_( hat(j))x^( hat(ȷ)))dx^( hat(del)^(2))-2(epsilon_(j hat(k))x^( hat(k))omega^( hat(ℓ)))dx^( hat(0))dx^( hat(3))],[(13.71)+delta_(j hat(jk))dx^( hat(3))dx^( hat(k))+O(|x^( hat(3))|^(2))dx^( hat(alpha))dx^( hat(beta)).]:}\begin{align*} d s^{2}= & -\left(1+2 a_{\hat{j}} x^{\hat{\jmath}}\right) d x^{\hat{\partial}^{2}}-2\left(\epsilon_{j \hat{k}} x^{\hat{k}} \omega^{\hat{\ell}}\right) d x^{\hat{0}} d x^{\hat{3}} \\ & +\delta_{j \hat{j k}} d x^{\hat{3}} d x^{\hat{k}}+O\left(\left|x^{\hat{3}}\right|^{2}\right) d x^{\hat{\alpha}} d x^{\hat{\beta}} . \tag{13.71} \end{align*}ds2=(1+2aj^xȷ^)dx^22(ϵjk^xk^ω^)dx0^dx3^(13.71)+δjjk^dx3^dxk^+O(|x3^|2)dxα^dxβ^.
Several features of this line element deserve notice, as follows.
(1) On the observer's world line P 0 ( τ ) P 0 ( τ ) P_(0)(tau)\mathscr{P}_{0}(\tau)P0(τ)-i.e., x j ^ = 0 d s 2 = η α β d x α ^ d x β ^ x j ^ = 0 d s 2 = η α β d x α ^ d x β ^ x^( hat(j))=0-ds^(2)=eta_(alpha beta)dx^( hat(alpha))dx^( hat(beta))x^{\hat{j}}=0-d s^{2}=\eta_{\alpha \beta} d x^{\hat{\alpha}} d x^{\hat{\beta}}xj^=0ds2=ηαβdxα^dxβ^.
(2) The observer's acceleration shows up in a correction term to g 0 ^ 0 ^ g 0 ^ 0 ^ g_( hat(0) hat(0))g_{\hat{0} \hat{0}}g0^0^,
(13.72a) δ g 0 ^ 0 ^ = 2 a x (13.72a) δ g 0 ^ 0 ^ = 2 a x {:(13.72a)deltag_( hat(0) hat(0))=-2a*x:}\begin{equation*} \delta g_{\hat{0} \hat{0}}=-2 a \cdot x \tag{13.72a} \end{equation*}(13.72a)δg0^0^=2ax
which is proportional to distance along the acceleration direction. For the flat-spacetime derivation of this correction term, see §6.6.
(3) The observer's rotation relative to inertial-guidance gyroscopes shows up in a correction term to g 0 ^ j ^ g 0 ^ j ^ g_( hat(0) hat(j))g_{\hat{0} \hat{j}}g0^j^, which can be rewritten in 3 -vector notation
(13.72b) δ g 0 ^ j ^ e j ^ = x × ω = + ω × x (13.72b) δ g 0 ^ j ^ e j ^ = x × ω = + ω × x {:(13.72b)deltag_( hat(0) hat(j))e_( hat(j))=-x xx omega=+omega xx x:}\begin{equation*} \delta g_{\hat{0} \hat{j}} \boldsymbol{e}_{\hat{j}}=-\boldsymbol{x} \times \omega=+\omega \times \boldsymbol{x} \tag{13.72b} \end{equation*}(13.72b)δg0^j^ej^=x×ω=+ω×x
(4) These first-order corrections to the line element are unaffected by spacetime curvature and contain no information about curvature. Only at second order, O ( | x j ^ | 2 ) O x j ^ 2 O(|x^( hat(j))|^(2))O\left(\left|x^{\hat{j}}\right|^{2}\right)O(|xj^|2), will curvature begin to show up.
(5) In the special case of zero acceleration and zero rotation ( a = ω = 0 ) ( a = ω = 0 ) (a=omega=0)(\boldsymbol{a}=\boldsymbol{\omega}=0)(a=ω=0), the observer's proper reference frame reduces to a local Lorentz frame ( g α ^ β = η α β g α ^ β = η α β g_( hat(alpha)beta)=eta_(alpha beta)g_{\hat{\alpha} \beta}=\eta_{\alpha \beta}gα^β=ηαβ, Γ α ˙ β ^ γ ^ = 0 Γ α ˙ β ^ γ ^ = 0 Gamma^(alpha^(˙))_( hat(beta) hat(gamma))=0\Gamma^{\dot{\alpha}}{ }_{\hat{\beta} \hat{\gamma}}=0Γα˙β^γ^=0 ) all along his geodesic world line! By contrast, the local Lorentz coordinate
Metric of proper reference frame, and its physical interpretation
systems constructed earlier in the book ("general" local Lorentz coordinates of §8.6, "Riemann normal coordinates" of §11.6) are local Lorentz only at a single event.
In the case of zero rotation and zero acceleration, one can derive the following expression for the metric, accurate to second order in | x j | x j |x^(j)|\left|x^{j}\right||xj| :
d s 2 = ( 1 R 0 ^ O ^ ^ m ^ x l ^ x m ^ ) d t 2 ( 4 3 R 0 ^ l ^ j ^ x x m ^ ) d t d x j ^ (13.73) + ( δ i j ^ 1 3 R i l ^ j ^ m ^ x ı ^ x m ^ ) d x ı ^ d x ȷ ^ + O ( | x j ^ | 3 ) d x α ^ d x β ^ d s 2 = 1 R 0 ^ O ^ ^ m ^ x l ^ x m ^ d t 2 4 3 R 0 ^ l ^ j ^ x x m ^ d t d x j ^ (13.73) + δ i j ^ 1 3 R i l ^ j ^ m ^ x ı ^ x m ^ d x ı ^ d x ȷ ^ + O x j ^ 3 d x α ^ d x β ^ {:[ds^(2)=(-1-R_( hat(0) hat(hat(O)) hat(m))x^( hat(l))x^( hat(m)))dt^(2)-((4)/(3)R_( hat(0) hat(l) hat(j))x^(ℓ)x^( hat(m)))dtdx^( hat(j))],[(13.73)+(delta_( hat(ij))-(1)/(3)R_(i hat(l) hat(j) hat(m))x^( hat(ı))x^( hat(m)))dx^( hat(ı))dx^( hat(ȷ))+O(|x^( hat(j))|^(3))dx^( hat(alpha))dx^( hat(beta))]:}\begin{align*} d s^{2}= & \left(-1-R_{\hat{0} \hat{\hat{O}} \hat{m}} x^{\hat{l}} x^{\hat{m}}\right) d t^{2}-\left(\frac{4}{3} R_{\hat{0} \hat{l} \hat{j}} x^{\ell} x^{\hat{m}}\right) d t d x^{\hat{j}} \\ & +\left(\delta_{\hat{i j}}-\frac{1}{3} R_{i \hat{l} \hat{j} \hat{m}} x^{\hat{\imath}} x^{\hat{m}}\right) d x^{\hat{\imath}} d x^{\hat{\jmath}}+O\left(\left|x^{\hat{j}}\right|^{3}\right) d x^{\hat{\alpha}} d x^{\hat{\beta}} \tag{13.73} \end{align*}ds2=(1R0^O^^m^xl^xm^)dt2(43R0^l^j^xxm^)dtdxj^(13.73)+(δij^13Ril^j^m^xı^xm^)dxı^dxȷ^+O(|xj^|3)dxα^dxβ^
[see, e.g., Manasse and Misner (1963)]. Here R a ^ β ^ γ ^ δ ^ R a ^ β ^ γ ^ δ ^ R_( hat(a) hat(beta) hat(gamma) hat(delta))R_{\hat{a} \hat{\beta} \hat{\gamma} \hat{\delta}}Ra^β^γ^δ^ are the components of the Riemann tensor along the world line x j = 0 x j = 0 x^(j)=0x^{j}=0xj=0. Such coordinates are called "Fermi Normal Coordinates."

EXERCISES

Exercise 13.14. INERTIAL AND CORIOLIS FORCES

An accelerated observer studies the path of a freely falling particle as it passes through the origin of his proper reference frame. If
(13.74) v ( d x 3 / d x 6 ) e 3 (13.74) v d x 3 / d x 6 e 3 {:(13.74)v-=(dx^(3)//dx^(6))e_(3):}\begin{equation*} \boldsymbol{v} \equiv\left(d x^{3} / d x^{6}\right) \boldsymbol{e}_{3} \tag{13.74} \end{equation*}(13.74)v(dx3/dx6)e3
is the particle's ordinary velocity, show that its ordinary acceleration relative to the observer's proper reference frame is
Here a a a\boldsymbol{a}a is the observer's own 4 -acceleration, and ω ω omega\omegaω is the angular velocity with which his spatial basis vectors e j e j e_(j)\boldsymbol{e}_{\boldsymbol{j}}ej are rotating [see equations (13.62)]. [Hint: Use the geodesic equation at the point x j = 0 x j = 0 x^(j)=0x^{j}=0xj=0 of the particle's trajectory. Note: This result was derived in flat spacetime in exercise 6.8 using a different method.]

Exercise 13.15. ROTATION GROUP: METRIC

(Continuation of exercises 9.13, 9.14, 10.17 and 11.12). Show that for the manifold SO ( 3 ) SO ( 3 ) SO(3)\mathrm{SO}(3)SO(3) of the rotation group, there exists a metric g g g\boldsymbol{g}g that is compatible with the covariant derivative grad\boldsymbol{\nabla}. Prove existence by exhibiting the metric components explicitly in the noncoordinate basis of generators { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα}. [Answer:
(13.76) g α β = δ α β . (13.76) g α β = δ α β . {:(13.76)g_(alpha beta)=delta_(alpha beta).:}\begin{equation*} g_{\alpha \beta}=\delta_{\alpha \beta} . \tag{13.76} \end{equation*}(13.76)gαβ=δαβ.
Restated in words: If one postulates that: (1) the manifold of the rotation group is locally Euclidean; (2) the generators of infinitesimal rotations { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} are orthonormal, e α e β = δ α β e α e β = δ α β e_(alpha)*e_(beta)=delta_(alpha beta)\boldsymbol{e}_{\alpha} \cdot \boldsymbol{e}_{\beta}=\delta_{\alpha \beta}eαeβ=δαβ; and (3) { e α } e α {e_(alpha)}\left\{\boldsymbol{e}_{\alpha}\right\}{eα} obey the standard rotation-group commutation relations
(13.77) [ e α , e β ] = ϵ α β γ e γ ; (13.77) e α , e β = ϵ α β γ e γ ; {:(13.77)[e_(alpha),e_(beta)]=-epsilon_(alpha beta gamma)e_(gamma);:}\begin{equation*} \left[\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right]=-\boldsymbol{\epsilon}_{\alpha \beta \gamma} \boldsymbol{e}_{\gamma} ; \tag{13.77} \end{equation*}(13.77)[eα,eβ]=ϵαβγeγ;
then the resulting geodesics of S O ( 3 ) S O ( 3 ) SO(3)S O(3)SO(3) agree with the geodesics chosen in exercise 10.17.]

снартев 14

CALCULATION OF CURVATURE

§14.1. CURVATURE AS A TOOL FOR UNDERSTANDING PHYSICS

Elementary physics sometimes allows one to shortcircuit any systematized calculation of curvature (frequency of oscillation of test particle; tide-producing acceleration near a center of attraction; curvature of a closed 3-sphere model universe; effect of parallel transport on gyroscope or vector; see Figures 1.1, 1.10, and 1.12, and Boxes 1.6 and 1.7); but on other occasions a calculation of curvature is the quickest way into the physics. This chapter is designed for such occasions. It describes three ways to calculate curvature and gives the components of the Einstein curvature tensor for a plane gravitational wave (Box 14.4, equation 5), for the Friedmann geometry of the universe (Box 14.5), and for Schwarzschild geometry, both static (exercise
This chapter is entirely Track 2.
Chapter 4 (differential forms) and Chapter 10, 11, and 13 (differential geometry) are necessary preparation for §§14.5-14.6.
This chapter is needed as preparation for Chapter 15 (Bianchi identities).
It will be helpful in many applications of gravitation theory (Chapters 23-40). 14.13) and dynamic (exercise 14.16). These and other calculations of curvature elsewhere are indexed under "curvature tensors."
It is enough to look at an expression for a 4-geometry as complicated as
d s 2 = ( x / 3 1 / 2 L + y 2 / 12 L 2 ) 3 1 / 2 ( v d z z ) 1 ( z / L ) 3 1 / 2 d t 2 + ( x / 3 1 / 2 L + y 2 / 12 L 2 ) 1 + 3 1 / 2 ( v d z z ) 1 + 2 / 3 1 / 2 ( z / L ) 1 + 3 1 / 2 d x 2 + ( x / 3 1 / 2 L + y 2 / 12 L 2 ) 2 + 3 1 / 2 ( v d z z ) 1 + 2 / 3 1 / 2 ( z / L ) 3 1 / 2 d y 2 + ( x / 3 1 / 2 L + y 2 / 12 L 2 ) 3 + 3 1 / 2 ( v d z z ) 1 + 2 / 3 1 / 2 ( z / L ) 2 3 1 / 2 × (14.1) × ( v 2 1 1 z / L ) d z 2 d s 2 = x / 3 1 / 2 L + y 2 / 12 L 2 3 1 / 2 v d z z 1 ( z / L ) 3 1 / 2 d t 2 + x / 3 1 / 2 L + y 2 / 12 L 2 1 + 3 1 / 2 v d z z 1 + 2 / 3 1 / 2 ( z / L ) 1 + 3 1 / 2 d x 2 + x / 3 1 / 2 L + y 2 / 12 L 2 2 + 3 1 / 2 v d z z 1 + 2 / 3 1 / 2 ( z / L ) 3 1 / 2 d y 2 + x / 3 1 / 2 L + y 2 / 12 L 2 3 + 3 1 / 2 v d z z 1 + 2 / 3 1 / 2 ( z / L ) 2 3 1 / 2 × (14.1) × v 2 1 1 z / L d z 2 {:[ds^(2)=-(x//3^(1//2)L+y^(2)//12L^(2))^(-3^(1//2))(int(vdz)/(z))^(-1)(-z//L)^(3^(-1//2))dt^(2)],[+(x//3^(1//2)L+y^(2)//12L^(2))^(1+3^(1//2))(int(vdz)/(z))^(1+2//3^(1//2))(-z//L)^(-1+3^(-1//2))dx^(2)],[+(x//3^(1//2)L+y^(2)//12L^(2))^(2+3^(1//2))(int(vdz)/(z))^(1+2//3^(1//2))(-z//L)^(-3^(-1//2))dy^(2)],[+(x//3^(1//2)L+y^(2)//12L^(2))^(3+3^(1//2))(int(vdz)/(z))^(1+2//3^(1//2))(-z//L)^(-2-3^(-1//2))xx],[(14.1) xx((v^(2)-1)/(-1-z//L))dz^(2)]:}\begin{align*} d s^{2}= & -\left(x / 3^{1 / 2} L+y^{2} / 12 L^{2}\right)^{-3^{1 / 2}}\left(\int \frac{v d z}{z}\right)^{-1}(-z / L)^{3^{-1 / 2}} d t^{2} \\ & +\left(x / 3^{1 / 2} L+y^{2} / 12 L^{2}\right)^{1+3^{1 / 2}}\left(\int \frac{v d z}{z}\right)^{1+2 / 3^{1 / 2}}(-z / L)^{-1+3^{-1 / 2}} d x^{2} \\ & +\left(x / 3^{1 / 2} L+y^{2} / 12 L^{2}\right)^{2+3^{1 / 2}}\left(\int \frac{v d z}{z}\right)^{1+2 / 3^{1 / 2}}(-z / L)^{-3^{-1 / 2}} d y^{2} \\ & +\left(x / 3^{1 / 2} L+y^{2} / 12 L^{2}\right)^{3+3^{1 / 2}}\left(\int \frac{v d z}{z}\right)^{1+2 / 3^{1 / 2}}(-z / L)^{-2-3^{-1 / 2}} \times \\ & \times\left(\frac{v^{2}-1}{-1-z / L}\right) d z^{2} \tag{14.1} \end{align*}ds2=(x/31/2L+y2/12L2)31/2(vdzz)1(z/L)31/2dt2+(x/31/2L+y2/12L2)1+31/2(vdzz)1+2/31/2(z/L)1+31/2dx2+(x/31/2L+y2/12L2)2+31/2(vdzz)1+2/31/2(z/L)31/2dy2+(x/31/2L+y2/12L2)3+31/2(vdzz)1+2/31/2(z/L)231/2×(14.1)×(v211z/L)dz2
Situations in which one must compute curvature
"Standard procedure" for computing curvature
Methods of displaying curvature formulas
Computation of curvature using a computer
[Harrison (1959)] to realize that one might understand the physical situation better if one knew what the curvature is; similarly with any other complicated expressions for metrics that arise from solving Einstein's equations or that appear undigested in the literature. In any such case, the appropriate method often is: curvature first, understanding second.
Curvature is the simplest local measure of geometric properties (see Box 14.1). Curvature is therefore a good first step toward a more comprehensive picture of the spacetime in question.
One sometimes has an expression for a spacetime metric first, and then makes calculations of curvature to understand it. But more often one makes calculations of curvature, subject to specified conditions of symmetry in space and time, as an aid in arriving at an expression for a physically interesting metric (stars, Chapters 23 to 26; model cosmologies, Chapters 27 to 30 ; collapse and black holes, Chapters 31 to 34 ; and gravitational waves, Chapters 35 to 37 ).
The basic "standard procedure for computing curvature" is illustrated in Box 14.2. Two formulas in Box 14.2 , derived previously, are used in succession. The first (equations 1 and 2) has the form Γ g g Γ g g Gamma∼g del g\Gamma \sim g \partial gΓgg and provides the Γ α β μ Γ α β μ Gamma_(alpha beta)^(mu)\Gamma_{\alpha \beta}^{\mu}Γαβμ. The other (equation 3) has the form R Γ + Γ 2 R Γ + Γ 2 R∼del Gamma+Gamma^(2)R \sim \partial \Gamma+\Gamma^{2}RΓ+Γ2 and gives the curvature components R μ ν α β R μ ν α β R^(mu)_(nu alpha beta)R^{\mu}{ }_{\nu \alpha \beta}Rμναβ.
After the curvature components have been computed, there are helpful ways to present the results. (1) Form the Ricci tensor R μ ν = R α μ α ν R μ ν = R α μ α ν R_(mu nu)=R^(alpha)_(mu alpha nu)R_{\mu \nu}=R^{\alpha}{ }_{\mu \alpha \nu}Rμν=Rαμαν and the scalar curvature R = R μ μ R = R μ μ R=R^(mu)_(mu^('))R=R^{\mu}{ }_{\mu^{\prime}}R=Rμμ (2) Form other invariants such as R μ ν α β R α β μ ν R μ ν α β R α β μ ν R^(mu nu)_(alpha beta)R^(alpha beta)_(mu nu)R^{\mu \nu}{ }_{\alpha \beta} R^{\alpha \beta}{ }_{\mu \nu}RμναβRαβμν. (3) Form components R μ ^ v ^ α ^ β ^ R μ ^ v ^ α ^ β ^ R^( hat(mu) hat(v))_( hat(alpha) hat(beta))R^{\hat{\mu} \hat{v}}{ }_{\hat{\alpha} \hat{\beta}}Rμ^v^α^β^ in a judiciously chosen orthonormal frame ω α ^ = L α ^ β d x β ω α ^ = L α ^ β d x β omega^( hat(alpha))=L^( hat(alpha))_(beta)dx^(beta)\boldsymbol{\omega}^{\hat{\alpha}}=L^{\hat{\alpha}}{ }_{\beta} \boldsymbol{d} x^{\beta}ωα^=Lα^βdxβ, and (4) display R [ μ ^ ν ^ ] [ α ^ β ^ ] R [ μ ^ ν ^ ] [ α ^ β ^ ] R^([ hat(mu) hat(nu)])_([ hat(alpha) hat(beta)])R^{[\hat{\mu} \hat{\nu}]}{ }_{[\hat{\alpha} \hat{\beta}]}R[μ^ν^][α^β^] as a 6 × 6 6 × 6 6xx66 \times 66×6 matrix (in four dimensions; a 3 × 3 3 × 3 3xx33 \times 33×3 matrix in three dimensions) where [ μ ^ ν ^ ] = [ 0 ^ 1 ^ ] , [ 0 ^ 2 ^ ] , [ 0 ^ 3 ^ ] , [ 2 2 ^ } , [ 3 ^ 1 ^ ] , [ 1 ^ 2 ^ ] [ μ ^ ν ^ ] = [ 0 ^ 1 ^ ] , [ 0 ^ 2 ^ ] , [ 0 ^ 3 ^ ] , [ 2 2 ^ } , [ 3 ^ 1 ^ ] , [ 1 ^ 2 ^ ] [ hat(mu) hat(nu)]=[ hat(0) hat(1)],[ hat(0) hat(2)],[ hat(0) hat(3)],[2 hat(2)},[ hat(3) hat(1)],[ hat(1) hat(2)][\hat{\mu} \hat{\nu}]=[\hat{0} \hat{1}],[\hat{0} \hat{2}],[\hat{0} \hat{3}],[2 \hat{2}\},[\hat{3} \hat{1}],[\hat{1} \hat{2}][μ^ν^]=[0^1^],[0^2^],[0^3^],[22^},[3^1^],[1^2^] labels the rows and [ α ^ β ^ ] [ α ^ β ^ ] [ hat(alpha) hat(beta)][\hat{\alpha} \hat{\beta}][α^β^] labels the columns (exercises 14.14 and 14.15). (5) Last, but by far the most important for general relativity, form the Einstein tensor G μ ^ ν ^ G μ ^ ν ^ G^( hat(mu))_( hat(nu))G^{\hat{\mu}}{ }_{\hat{\nu}}Gμ^ν^ as described in §14.2.
The method of computation outlined above and described in more detail in Box 14.2 is used wherever it is quicker to employ a standard method than to learn or invent a better method. The standard method is always preferable for the student in a short course where physical insight has higher priority than technical facility. It is, however, a dull method, better suited to computers than to people. Even the algebra can be handled by a computer (see Box 14.3).

EXERCISES

Exercise 14.1. CURVATURE OF A TWO-DIMENSIONAL HYPERBOLOID

Compute the curvature of the hyperboloid t 2 x 2 y 2 = T 2 = t 2 x 2 y 2 = T 2 = t^(2)-x^(2)-y^(2)=T^(2)=t^{2}-x^{2}-y^{2}=T^{2}=t2x2y2=T2= const in 2 + 1 2 + 1 2+12+12+1 Minkowski spacetime with d s 3 2 = d t 2 + d x 2 + d y 2 d s 3 2 = d t 2 + d x 2 + d y 2 ds_(3)^(2)=-dt^(2)+dx^(2)+dy^(2)d s_{3}{ }^{2}=-d t^{2}+d x^{2}+d y^{2}ds32=dt2+dx2+dy2. First show that intervals within this two-dimensional surface can be expressed in the form d s 2 = T 2 ( d α 2 + sinh 2 α d ϕ 2 ) d s 2 = T 2 d α 2 + sinh 2 α d ϕ 2 ds^(2)=T^(2)(dalpha^(2)+sinh^(2)alpha dphi^(2))d s^{2}=T^{2}\left(d \alpha^{2}+\sinh ^{2} \alpha d \phi^{2}\right)ds2=T2(dα2+sinh2αdϕ2) by a suitable choice of coordinates α , ϕ α , ϕ alpha,phi\alpha, \phiα,ϕ, on the hyperboloid.
Exercise 14.2. RIEMANNIAN CURVATURE EXPRESSIBLE IN TERMS OF
RICCI CURVATURE IN TWO AND THREE DIMENSIONS
In two dimensions, there is only one independent curvature component, R 1212 R 1212 R_(1212)R_{1212}R1212. Evidently the single scalar quantity R R RRR must carry the same information. The two-dimensional identity R μ ν α β = 1 2 R ( g μ α g ν β g μ β g ν α ) R μ ν α β = 1 2 R g μ α g ν β g μ β g ν α R_(mu nu alpha beta)=(1)/(2)R(g_(mu alpha)g_(nu beta)-g_(mu beta)g_(nu alpha))R_{\mu \nu \alpha \beta}=\frac{1}{2} R\left(g_{\mu \alpha} g_{\nu \beta}-g_{\mu \beta} g_{\nu \alpha}\right)Rμναβ=12R(gμαgνβgμβgνα) is established by noting that it is the only tensor formula giving
(continued on page 343)

Box 14.1 PERSPECTIVES ON CURVATURE

  1. Historical point of departure: a curved line on a plane. There is no way to define the curvature of a line by measurements confined to ("intrinsic to") the line itself. One needs, for example, the azimuthal bearing θ θ theta\thetaθ of the tangent vector relative to a fixed direction in the plane, as a function of proper distance s s sss measured along the curve; thus, θ = θ ( s ) θ = θ ( s ) theta=theta(s)\theta=\theta(s)θ=θ(s). Then curvature κ κ kappa\kappaκ and its reciprocal, the radius of curvature ρ ρ rho\rhoρ, are given by κ ( s ) = κ ( s ) = kappa(s)=\kappa(s)=κ(s)= 1 / ρ ( s ) = d θ ( s ) / d s 1 / ρ ( s ) = d θ ( s ) / d s 1//rho(s)=d theta(s)//ds1 / \rho(s)=d \theta(s) / d s1/ρ(s)=dθ(s)/ds. Alternatively, one can examine departure, y y yyy, measured normally off from the tan-
    gent line as a function of distance x x xxx measured along that tangent line; then κ = 1 / ρ = d 2 y / d x 2 κ = 1 / ρ = d 2 y / d x 2 kappa=1//rho=d^(2)y//dx^(2)\kappa=1 / \rho=d^{2} y / d x^{2}κ=1/ρ=d2y/dx2.
  2. This concept was later extended to a curved surface embedded in flat (Euclidean) 3 -space. Departure, z z zzz, of the smooth curved surface from the flat surface tangent to it at a given point is described in the neighborhood of that point by the quadratic expression
z = 1 2 a x 2 + b x y + 1 2 c y 2 . z = 1 2 a x 2 + b x y + 1 2 c y 2 . z=(1)/(2)ax^(2)+bxy+(1)/(2)cy^(2).z=\frac{1}{2} a x^{2}+b x y+\frac{1}{2} c y^{2} .z=12ax2+bxy+12cy2.
Rotation of the axes by an appropriate angle α α alpha\alphaα,
x = ξ cos α + η sin α , y = ξ sin α + η cos α , x = ξ cos α + η sin α , y = ξ sin α + η cos α , {:[x=xi cos alpha+eta sin alpha","],[y=-xi sin alpha+eta cos alpha","]:}\begin{aligned} & x=\xi \cos \alpha+\eta \sin \alpha, \\ & y=-\xi \sin \alpha+\eta \cos \alpha, \end{aligned}x=ξcosα+ηsinα,y=ξsinα+ηcosα,
reduces this expression to
z = 1 2 κ 1 ξ 2 + 1 2 κ 2 η 2 , z = 1 2 κ 1 ξ 2 + 1 2 κ 2 η 2 , z=(1)/(2)kappa_(1)xi^(2)+(1)/(2)kappa_(2)eta^(2),z=\frac{1}{2} \kappa_{1} \xi^{2}+\frac{1}{2} \kappa_{2} \eta^{2},z=12κ1ξ2+12κ2η2,
with
κ 1 = 1 / ρ 1 κ 1 = 1 / ρ 1 kappa_(1)=1//rho_(1)\kappa_{1}=1 / \rho_{1}κ1=1/ρ1
and
κ 2 = 1 / ρ 2 κ 2 = 1 / ρ 2 kappa_(2)=1//rho_(2)\kappa_{2}=1 / \rho_{2}κ2=1/ρ2
representing the two "principal curvatures" of the surface.
3. Gauss (1827) conceived the idea of defining curvature by measurements confined entirely to the surface ("society of ants"). From a given point P P P\mathscr{P}P on the surface, proceed on a geodesic on the surface for a proper distance ϵ ϵ epsilon\epsilonϵ measured entirely within the surface. Repeat, starting at the original point but proceeding in other directions.

Box 14.1 (continued)

Obtain an infinity of points. They define a "circle". Determine its proper circumference, again by measurements confined entirely to the surface. Using the metric corresponding to the embedding viewpoint
d s 2 = d z 2 + d ξ 2 + d η 2 (Euclidean 3-space) = [ ( κ 1 ξ d ξ + κ 2 η d η ) 2 + ( d ξ 2 + d η 2 ) ] ( metric intrinsic to the curved 2 -geometry ) , d s 2 = d z 2 + d ξ 2 + d η 2  (Euclidean 3-space)  = κ 1 ξ d ξ + κ 2 η d η 2 + d ξ 2 + d η 2  metric intrinsic   to the curved  2 -geometry  , {:[ds^(2)=dz^(2)+dxi^(2)+deta^(2)quad" (Euclidean 3-space) "],[=[(kappa_(1)xi d xi+kappa_(2)eta d eta)^(2)+(dxi^(2)+deta^(2))]quad([" metric intrinsic "],[" to the curved "],[2"-geometry "])","]:}\begin{aligned} d s^{2} & =d z^{2}+d \xi^{2}+d \eta^{2} \quad \text { (Euclidean 3-space) } \\ & =\left[\left(\kappa_{1} \xi d \xi+\kappa_{2} \eta d \eta\right)^{2}+\left(d \xi^{2}+d \eta^{2}\right)\right] \quad\left(\begin{array}{l} \text { metric intrinsic } \\ \text { to the curved } \\ 2 \text {-geometry } \end{array}\right), \end{aligned}ds2=dz2+dξ2+dη2 (Euclidean 3-space) =[(κ1ξdξ+κ2ηdη)2+(dξ2+dη2)]( metric intrinsic  to the curved 2-geometry ),
one can calculate the result of such an "intrinsic measurement." One calculates that the circumference differs from the Euclidean value, 2 π ϵ 2 π ϵ 2pi epsilon2 \pi \epsilon2πϵ, by a fractional correction that is proportional to the square of ϵ ϵ epsilon\epsilonϵ; specifically,
Lim ϵ 0 6 ϵ 2 ( 1 circumference 2 π ϵ ) = κ 1 κ 2 = 1 ρ 1 ρ 2 = det ( a b b c ) Lim ϵ 0 6 ϵ 2 1  circumference  2 π ϵ = κ 1 κ 2 = 1 ρ 1 ρ 2 = det a b b c Lim_(epsilon rarr0)(6)/(epsilon^(2))(1-(" circumference ")/(2pi epsilon))=kappa_(1)kappa_(2)=(1)/(rho_(1)rho_(2))=det([a,b],[b,c])\operatorname{Lim}_{\epsilon \rightarrow 0} \frac{6}{\epsilon^{2}}\left(1-\frac{\text { circumference }}{2 \pi \epsilon}\right)=\kappa_{1} \kappa_{2}=\frac{1}{\rho_{1} \rho_{2}}=\operatorname{det}\left(\begin{array}{ll} a & b \\ b & c \end{array}\right)Limϵ06ϵ2(1 circumference 2πϵ)=κ1κ2=1ρ1ρ2=det(abbc)
Note especially the first equality sign. Gauss did not conceal the elation he felt on discovering that something defined by measurements entirely within the surface agrees with the product of two quantities, κ 1 κ 1 kappa_(1)\kappa_{1}κ1 and κ 2 κ 2 kappa_(2)\kappa_{2}κ2, that individually demand for their definition measurements extrinsic to the surface.
4. The contrast between "extrinsic" and "intrinsic" curvature is summarized in the terms,
( extrinsic curvature ) = κ = ( κ 1 + κ 2 ) ( cm 1 ) ( intrinsic or Gaussian curvature ) = κ 1 κ 2 ( cm 2 ) (  extrinsic curvature  ) = κ = κ 1 + κ 2 cm 1 (  intrinsic or Gaussian   curvature  ) = κ 1 κ 2 cm 2 {:[(" extrinsic curvature ")=kappa=(kappa_(1)+kappa_(2))(cm^(-1))],[((" intrinsic or Gaussian ")/(" curvature "))=kappa_(1)kappa_(2)(cm^(-2))]:}\begin{gathered} (\text { extrinsic curvature })=\kappa=\left(\kappa_{1}+\kappa_{2}\right)\left(\mathrm{cm}^{-1}\right) \\ \binom{\text { intrinsic or Gaussian }}{\text { curvature }}=\kappa_{1} \kappa_{2}\left(\mathrm{~cm}^{-2}\right) \end{gathered}( extrinsic curvature )=κ=(κ1+κ2)(cm1)( intrinsic or Gaussian  curvature )=κ1κ2( cm2)
(the latter being identical with half the scalar curvature invariant, R R RRR, of the 2 -geometry). Draw a 3 : 4 : 5 3 : 4 : 5 3:4:53: 4: 53:4:5 triangle on a flat piece of paper; then curl up the paper. The Euclidean 2 -geometry intrinsic to the piece of paper is preserved by this bending.
The Gaussian curvature intrinsic to the surface remains unaltered; it keeps the Euclidean value of zero ( κ 2 κ 2 kappa_(2)\kappa_{2}κ2, non-zero; κ 1 κ 1 kappa_(1)\kappa_{1}κ1, zero; product, κ 1 κ 2 = κ 1 κ 2 = kappa_(1)kappa_(2)=\kappa_{1} \kappa_{2}=κ1κ2= zero). However, the extrinsic curvature is changed from κ 1 + κ 2 = 0 κ 1 + κ 2 = 0 kappa_(1)+kappa_(2)=0\kappa_{1}+\kappa_{2}=0κ1+κ2=0 to a non-zero value, κ 1 + κ 1 + kappa_(1)+\kappa_{1}+κ1+ κ 2 0 κ 2 0 kappa_(2)!=0\kappa_{2} \neq 0κ20.

5. The curvature dealt with in this chapter is curvature intrinsic to spacetime; that is, curvature defined without any use of, and repelling every thought of, any embedding in any hypothetical higher-dimensional flat manifold (concept of Riemann,
Clifford, and Einstein that geometry is a dynamic participant in physics, not some God-given perfection above the battles of matter and energy).
6. The curvature of the geometry of spacetime imposes curvature on any spacelike slice (3-geometry; "initial-value hypersurface") through that spacetime (see "relations of Gauss and Codazzi" in Chapter 21, on the initial-value problem of geometrodynamics).
7. Rotation of a vector transported parallel to itself around a closed loop provides a definition of curvature as useful in four and three as in two dimensions. (In a curved two-dimensional geometry, at a point there is only one plane. Consequently only one number is required to describe the Gaussian curvature there. In three and four dimensions, there are more independent planes through a point and therefore more numbers are required to describe the curvature.) In the diagram, start with a vector at position 1 (North Pole). Transport it parallel to itself (position 2,3, ...) around a 90 90 90 90 90 90 90^(@)-90^(@)-90^(@)90^{\circ}-90^{\circ}-90^{\circ}909090 spherical triangle. It arrives back at the starting point (position 4) turned through 90 90 90^(@)90^{\circ}90 :

( Gaussian curvature ) = ( angle turned through ) ( area circum- navigated ) = ( π / 2 ) ( 1 / 8 ) ( 4 π a 2 ) = 1 a 2 (  Gaussian   curvature  ) = (  angle turned   through  ) (  area circum-   navigated  ) = ( π / 2 ) ( 1 / 8 ) 4 π a 2 = 1 a 2 ((" Gaussian ")/(" curvature "))=(((" angle turned ")/(" through ")))/(((" area circum- ")/(" navigated ")))=((pi//2))/((1//8)(4pia^(2)))=(1)/(a^(2))\binom{\text { Gaussian }}{\text { curvature }}=\frac{\binom{\text { angle turned }}{\text { through }}}{\binom{\text { area circum- }}{\text { navigated }}}=\frac{(\pi / 2)}{(1 / 8)\left(4 \pi a^{2}\right)}=\frac{1}{a^{2}}( Gaussian  curvature )=( angle turned  through )( area circum-  navigated )=(π/2)(1/8)(4πa2)=1a2
(positive; sense of rotation same as sense of circumnavigation).
8. Still staying for simplicity with a curved twodimensional manifold, describe the curvature of the 2 -surface as a 2 -form ("box-like structure") defined over the entire surface. The number of boxes enclosed by any given route gives immediately the angle in radians (or tenths or hundredths of a radian, etc., depending on chosen fineness of subdivision) turned through by a vector carried parallel to itself around that route. The contribution of a given box is counted as positive or negative depending on whether the sense of the arrow marked on it (see magnified view) agrees or disagrees with the sense of circumnavigation of the route.
Box 14.1 (continued)

9. Curvature 2 -form for the illustrated surface of rotational symmetry ("pith helmet") with metric d s 2 = d σ 2 + r 2 ( σ ) d ϕ 2 d s 2 = d σ 2 + r 2 ( σ ) d ϕ 2 ds^(2)=dsigma^(2)+r^(2)(sigma)dphi^(2)d s^{2}=d \sigma^{2}+r^{2}(\sigma) d \phi^{2}ds2=dσ2+r2(σ)dϕ2 is
(1) curvature = 1 r d 2 r d σ 2 d σ r d ϕ (1)  curvature  = 1 r d 2 r d σ 2 d σ r d ϕ {:(1)" curvature "=-(1)/(r)(d^(2)r)/(dsigma^(2))d sigma^^rd phi:}\begin{equation*} \text { curvature }=-\frac{1}{r} \frac{d^{2} r}{d \sigma^{2}} \boldsymbol{d} \sigma \wedge r \boldsymbol{d} \phi \tag{1} \end{equation*}(1) curvature =1rd2rdσ2dσrdϕ
(positive on crown of helmet, negative around brim, as indicated by sense of arrows in the "boxes of the 2 -form" shown at left). "Meaning" of r r rrr is illustrated by imbedding the surface in Euclidean 3-space, a convenience for visualization; but more important is the idea of a 2-geometry defined by measurements intrinsic to it, with no embedding.
10. How lengths ("metric") determine curvature in quantitative detail is shown nowhere more clearly than in this two-dimensional example, a model for "what is going on behind the scene" in the mathematical calculations done in this chapter with 1 -forms and 2 -forms in four-dimensional spacetime.
a. Net rotation in going around element of surface C B B B C ¯ C C B B B C ¯ C CBBB bar(C)C\mathscr{C B B} \mathscr{B} \bar{C} \mathscr{C}CBBBC¯C is δ δ ¯ δ δ ¯ delta- bar(delta)\delta-\bar{\delta}δδ¯ (no turn of vector to left or to right in its transport along a meridian a a ¯ a a ¯ a bar(a)a \bar{a}aa¯ or G B ¯ B ¯ ) G B ¯ B ¯ ) G bar(B) bar(B))\mathscr{G} \bar{B} \bar{B})GB¯B¯).
b. Rotation of vector in going from C C C\mathscr{C}C to B B B\mathscr{B}B, relative to coordinate system (directions of meridians), is
( angle δ ) = arc length = r ( σ + d σ ) Δ ϕ r ( σ ) Δ ϕ d σ = ( d r d σ ) σ Δ ϕ ˙ . (  angle  δ ) = arc  length  = r ( σ + d σ ) Δ ϕ r ( σ ) Δ ϕ d σ = d r d σ σ Δ ϕ ˙ (" angle "delta)=(arc)/(" length ")=(r(sigma+d sigma)Delta phi-r(sigma)Delta phi)/(d sigma)=((dr)/(d sigma))_(sigma)Deltaphi^(˙)". "(\text { angle } \delta)=\frac{\operatorname{arc}}{\text { length }}=\frac{r(\sigma+d \sigma) \Delta \phi-r(\sigma) \Delta \phi}{d \sigma}=\left(\frac{d r}{d \sigma}\right)_{\sigma} \Delta \dot{\phi} \text {. }( angle δ)=arc length =r(σ+dσ)Δϕr(σ)Δϕdσ=(drdσ)σΔϕ˙
c. Rotation of vector in going from a ¯ a ¯ bar(a)\bar{a}a¯ to B B ¯ bar(B)\overline{\mathscr{B}}B is similarly
(angle δ ¯ ) = ( d r d σ ) σ + Δ σ Δ ϕ  (angle  δ ¯ ) = d r d σ σ + Δ σ Δ ϕ " (angle " bar(delta))=((dr)/(d sigma))_(sigma+Delta sigma)Delta phi\text { (angle } \bar{\delta})=\left(\frac{d r}{d \sigma}\right)_{\sigma+\Delta \sigma} \Delta \phi (angle δ¯)=(drdσ)σ+ΔσΔϕ
d. Thus net rotation is:
δ δ ¯ = ( d 2 r d σ 2 ) σ Δ σ Δ φ ˙ δ δ ¯ = d 2 r d σ 2 σ Δ σ Δ φ ˙ delta- bar(delta)=-((d^(2)r)/(dsigma^(2)))_(sigma)Delta sigma Deltavarphi^(˙)\delta-\bar{\delta}=-\left(\frac{d^{2} r}{d \sigma^{2}}\right)_{\sigma} \Delta \sigma \Delta \dot{\varphi}δδ¯=(d2rdσ2)σΔσΔφ˙
e. Expressed as a form, this gives immediately equation (1).
f. Ideas and calculations are more complicated in four dimensions, primarily because one has to deal with different choices for the orientation of the surface to be studied at the point in question.
11. Translation of these geometric ideas into the language of forms is most immediate when one stays with this example of two dimensions. A sample vector A i = ( A 1 , A 2 ) A i = A 1 , A 2 A^(i)=(A^(1),A^(2))A^{i}=\left(A^{1}, A^{2}\right)Ai=(A1,A2) carried around the boundary of an element of surface comes back to its starting point slightly changed in direction:
(2) ( change in A i ) = R i A j . (2) (  change   in  A i ) = R i A j . {:(2)-((" change ")/(" in "A^(i)))=R^(i)A^(j).:}\begin{equation*} -\binom{\text { change }}{\text { in } A^{i}}=\mathscr{R}^{i} A^{j} . \tag{2} \end{equation*}(2)( change  in Ai)=RiAj.
a. To be more specific, it is convenient to adopt as the basis 1 -forms ω 1 ^ = d σ ω 1 ^ = d σ omega^( hat(1))=d sigma\boldsymbol{\omega}^{\hat{1}}=\boldsymbol{d} \boldsymbol{\sigma}ω1^=dσ and ω 2 ^ = r d ϕ ω 2 ^ = r d ϕ omega^( hat(2))=rd phi\boldsymbol{\omega}^{\hat{2}}=r \boldsymbol{d} \phiω2^=rdϕ, and have A 1 ^ A 1 ^ A^( hat(1))A^{\hat{1}}A1^ as the component of A A A\boldsymbol{A}A along the direction of increasing σ , A 2 ^ σ , A 2 ^ sigma,A^( hat(2))\sigma, A^{\hat{2}}σ,A2^ as the component of A A A\boldsymbol{A}A along the direction of increasing ϕ ϕ phi\phiϕ. The matrix R i ^ j R i ^ j R hat(i)_(j)\mathscr{R} \hat{i}_{j}Ri^j is a rotation matrix, which produces a change in direction but no change in length (zero diagonal components); thus here
(3) R j = 0 R 1 ^ i ^ r ^ 2 ^ 0 . (3) R j = 0 R 1 ^ i ^ r ^ 2 ^ 0 . {:(3)||R_(j)||=||[0,R hat(1)_( hat(i))],[-ℜ hat(r)_( hat(2)),0]||.:}\left\|\mathscr{R}{ }_{j}\right\|=\left\|\begin{array}{cc} 0 & \mathscr{R} \hat{1}_{\hat{i}} \tag{3}\\ -\Re \hat{r}_{\hat{2}} & 0 \end{array}\right\| .(3)Rj=0R1^i^r^2^0.
In this example, R 1 2 R 1 2 R_(1)_(2)\mathscr{R}{ }_{1}{ }_{2}R12 evidently represents the angle through which the vector A A A\boldsymbol{A}A turns on transport parallel to itself around the element of surface.
b. So far the rotation is "indefinite" because the size of the element of surface has not yet been specified. It is most conveniently conceived as an elementary parallelogram, defined by two vectors ("bivector"). Thus R i j ^ R i j ^ R^(i)_( hat(j))\mathscr{R}{ }^{i}{ }_{\hat{j}}Rij^, or, specifically, the one element that counts, R 1 2 R 1 2 R^(1_(2))\mathscr{R}^{1_{2}}R12 (the "angle of rotation"), has to be envisaged as a mathematical object (" 2 -form") endowed with two slots, into which these two vectors are inserted to get a definite number (angle in radians). In the example of the pith helmet, one has, from equation (1)
(4) R 2 ^ = 1 r d 2 r d σ 2 ω 1 ^ ω 2 ^ . (4) R 2 ^ = 1 r d 2 r d σ 2 ω 1 ^ ω 2 ^ . {:(4)R_( hat(2))=-(1)/(r)(d^(2)r)/(dsigma^(2))omega^( hat(1))^^omega^( hat(2)).:}\begin{equation*} \mathscr{R}_{\hat{2}}=-\frac{1}{r} \frac{d^{2} r}{d \sigma^{2}} \boldsymbol{\omega}^{\hat{1}} \wedge \boldsymbol{\omega}^{\hat{2}} . \tag{4} \end{equation*}(4)R2^=1rd2rdσ2ω1^ω2^.
Thus the R μ ν R μ ν R^(mu)_(nu)\mathscr{R}^{\mu}{ }_{\nu}Rμν in the text are called "curvature 2 -forms."

Box 14.1 (continued)

c. The text tells one how to read out of such expressions the components of the Riemann curvature tensor; for example here,
R 1 ^ 2 ^ 1 ^ 2 ^ = R 1 ^ 2 2 ^ 1 ^ = ( 1 / r ) ( d 2 r / d σ 2 ) ( coefficients of ω 1 ^ ω 2 ^ or ω 2 ^ ω 1 ^ ) . R 1 ^ 2 ^ 1 ^ 2 ^ = R 1 ^ 2 2 ^ 1 ^ = ( 1 / r ) d 2 r / d σ 2  coefficients of  ω 1 ^ ω 2 ^  or  ω 2 ^ ω 1 ^ . R^( hat(1)_( hat(2) hat(1) hat(2)))=-R^( hat(1))_(2 hat(2) hat(1))=(-1//r)(d^(2)r//dsigma^(2))(" coefficients of "omega^( hat(1))^^omega^( hat(2))" or "omega^( hat(2))^^omega^( hat(1))).R^{\hat{1}_{\hat{2} \hat{1} \hat{2}}}=-R^{\hat{1}}{ }_{2 \hat{2} \hat{1}}=(-1 / r)\left(d^{2} r / d \sigma^{2}\right)\left(\text { coefficients of } \boldsymbol{\omega}^{\hat{1}} \wedge \boldsymbol{\omega}^{\hat{2}} \text { or } \boldsymbol{\omega}^{\hat{2}} \wedge \boldsymbol{\omega}^{\hat{1}}\right) .R1^2^1^2^=R1^22^1^=(1/r)(d2r/dσ2)( coefficients of ω1^ω2^ or ω2^ω1^).
d. Generalizing to four dimensions, one understands by R α β μ ν R α β μ ν R^(alpha)_(beta mu nu)R^{\alpha}{ }_{\beta \mu \nu}Rαβμν the factor that one has to multiply by three numbers to obtain a fourth. The number obtained is the change (with reversed sign) that takes place in the α α alpha\alphaα th component of a vector when that vector is transported parallel to itself around a closed path, defined, for example, by a parallelogram built from two vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v. The factors that multiply R α β μ ν R α β μ ν R^(alpha)_(beta mu nu)R^{\alpha}{ }_{\beta \mu \nu}Rαβμν are (1) the component of the vector A A A\boldsymbol{A}A in the β β beta\betaβ th direction and ( 2 , 3 ) ( 2 , 3 ) (2,3)(2,3)(2,3) the μ ν μ ν mu nu\mu \nuμν component of the extension of the parallelogram, ( u μ v ν u v v μ ) u μ v ν u v v μ (u^(mu)v^(nu)-u^(v)v^(mu))\left(u^{\mu} v^{\nu}-u^{v} v^{\mu}\right)(uμvνuvvμ). Thus
δ A α = R α β | μ v | A β ( u μ v ν u ν v μ ) δ A α = R α β | μ v | A β u μ v ν u ν v μ deltaA^(alpha)=-R^(alpha)_(beta|mu v|)A^(beta)(u^(mu)v^(nu)-u^(nu)v^(mu))\delta A^{\alpha}=-R^{\alpha}{ }_{\beta|\mu v|} A^{\beta}\left(u^{\mu} v^{\nu}-u^{\nu} v^{\mu}\right)δAα=Rαβ|μv|Aβ(uμvνuνvμ)

Box 14.2 STRAIGHTFORWARD CURVATURE COMPUTATION (Illustrated for a Globe)

The elementary and universally applicable method for computing the components R μ ν α β R μ ν α β R^(mu)_(nu alpha beta)R^{\mu}{ }_{\nu \alpha \beta}Rμναβ of the Riemann curvature tensor starts from the metric components g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν in a coordinate basis, and proceeds by the following scheme:
g μ ν Γ g Γ μ α β Γ α β μ R Γ + Γ Γ R μ ν α β . g μ ν Γ g Γ μ α β Γ α β μ R Γ + Γ Γ R μ ν α β . g_(mu nu)rarr"Gamma~~del g"Gamma_(mu alpha beta)longrightarrowGamma_(alpha beta)^(mu)rarr"R∼del Gamma+Gamma Gamma"R^(mu)_(nu alpha beta).g_{\mu \nu} \xrightarrow{\Gamma \approx \partial g} \Gamma_{\mu \alpha \beta} \longrightarrow \Gamma_{\alpha \beta}^{\mu} \xrightarrow{R \sim \partial \Gamma+\Gamma \Gamma} R^{\mu}{ }_{\nu \alpha \beta} .gμνΓgΓμαβΓαβμRΓ+ΓΓRμναβ.
The formulas required for these three steps are
(1) Γ μ α β = 1 2 ( g μ α x β + g μ β x α g α β x μ ) , (2) Γ α β μ = g μ ν Γ ν α β , (1) Γ μ α β = 1 2 g μ α x β + g μ β x α g α β x μ , (2) Γ α β μ = g μ ν Γ ν α β , {:[(1)Gamma_(mu alpha beta)=(1)/(2)((delg_(mu alpha))/(delx^(beta))+(delg_(mu beta))/(delx^(alpha))-(delg_(alpha beta))/(delx^(mu)))","],[(2)Gamma_(alpha beta)^(mu)=g^(mu nu)Gamma_(nu alpha beta)","]:}\begin{align*} \Gamma_{\mu \alpha \beta} & =\frac{1}{2}\left(\frac{\partial g_{\mu \alpha}}{\partial x^{\beta}}+\frac{\partial g_{\mu \beta}}{\partial x^{\alpha}}-\frac{\partial g_{\alpha \beta}}{\partial x^{\mu}}\right), \tag{1}\\ \Gamma_{\alpha \beta}^{\mu} & =g^{\mu \nu} \Gamma_{\nu \alpha \beta}, \tag{2} \end{align*}(1)Γμαβ=12(gμαxβ+gμβxαgαβxμ),(2)Γαβμ=gμνΓναβ,
and
(3) R μ ν α β = Γ μ ν β x α Γ μ v α x β + Γ p α μ Γ ν β ρ Γ p β μ Γ ρ v α (3) R μ ν α β = Γ μ ν β x α Γ μ v α x β + Γ p α μ Γ ν β ρ Γ p β μ Γ ρ v α {:(3)R^(mu)_(nu alpha beta)=(delGamma^(mu)_(nu beta))/(delx^(alpha))-(delGamma^(mu)_(v alpha))/(delx^(beta))+Gamma_(p alpha)^(mu)Gamma_(nu beta)^(rho)-Gamma_(p beta)^(mu)Gamma^(rho)_(v alpha):}\begin{equation*} R^{\mu}{ }_{\nu \alpha \beta}=\frac{\partial \Gamma^{\mu}{ }_{\nu \beta}}{\partial x^{\alpha}}-\frac{\partial \Gamma^{\mu}{ }_{v \alpha}}{\partial x^{\beta}}+\Gamma_{p \alpha}^{\mu} \Gamma_{\nu \beta}^{\rho}-\Gamma_{p \beta}^{\mu} \Gamma^{\rho}{ }_{v \alpha} \tag{3} \end{equation*}(3)Rμναβ=ΓμνβxαΓμvαxβ+ΓpαμΓνβρΓpβμΓρvα
The metric of the two-dimensional surface of a sphere of radius a a aaa is
(4) d s 2 = a 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (4) d s 2 = a 2 d θ 2 + sin 2 θ d ϕ 2 {:(4)ds^(2)=a^(2)(dtheta^(2)+sin^(2)theta dphi^(2)):}\begin{equation*} d s^{2}=a^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) \tag{4} \end{equation*}(4)ds2=a2(dθ2+sin2θdϕ2)
To compute the curvature by the standard method, use the formula for d s 2 d s 2 ds^(2)d s^{2}ds2 as a table of g k g k g_(kℓ)g_{k \ell}gk values. It shows that g θ θ = a 2 , g θ ϕ = 0 , g ϕ ϕ = a 2 sin 2 θ g θ θ = a 2 , g θ ϕ = 0 , g ϕ ϕ = a 2 sin 2 θ g_(theta theta)=a^(2),g_(theta phi)=0,g_(phi phi)=a^(2)sin^(2)thetag_{\theta \theta}=a^{2}, g_{\theta \phi}=0, g_{\phi \phi}=a^{2} \sin ^{2} \thetagθθ=a2,gθϕ=0,gϕϕ=a2sin2θ. Compute the six possible different Γ j k l = Γ j l k Γ j k l = Γ j l k Gamma_(jkl)=Gamma_(jlk)\Gamma_{j k l}=\Gamma_{j l k}Γjkl=Γjlk (there will be 40 in four dimensions) from formula
(1). Thus
Γ θ ϕ ϕ = a 2 sin θ cos θ = Γ ϕ ϕ θ , (5) Γ θ θ θ = Γ ϕ ϕ ϕ = 0 , Γ θ θ ϕ = Γ ϕ θ θ = 0 . Γ θ ϕ ϕ = a 2 sin θ cos θ = Γ ϕ ϕ θ , (5) Γ θ θ θ = Γ ϕ ϕ ϕ = 0 , Γ θ θ ϕ = Γ ϕ θ θ = 0 . {:[Gamma_(theta phi phi)=-a^(2)sin theta cos theta=-Gamma_(phi phi theta)","],[(5)Gamma_(theta theta theta)=Gamma_(phi phi phi)=0","],[Gamma_(theta theta phi)=Gamma_(phi theta theta)=0.]:}\begin{align*} \Gamma_{\theta \phi \phi} & =-a^{2} \sin \theta \cos \theta=-\Gamma_{\phi \phi \theta}, \\ \Gamma_{\theta \theta \theta} & =\Gamma_{\phi \phi \phi}=0, \tag{5}\\ \Gamma_{\theta \theta \phi} & =\Gamma_{\phi \theta \theta}=0 . \end{align*}Γθϕϕ=a2sinθcosθ=Γϕϕθ,(5)Γθθθ=Γϕϕϕ=0,Γθθϕ=Γϕθθ=0.
Raise the first index:
Γ ϕ ϕ θ = sin θ cos θ (6) Γ ϕ θ ϕ = cot θ Γ θ θ θ = Γ θ θ ϕ = 0 = Γ ϕ θ θ = Γ ϕ ϕ ϕ . Γ ϕ ϕ θ = sin θ cos θ (6) Γ ϕ θ ϕ = cot θ Γ θ θ θ = Γ θ θ ϕ = 0 = Γ ϕ θ θ = Γ ϕ ϕ ϕ . {:[Gamma_(phi phi)^(theta)=-sin theta cos theta],[(6)Gamma_(phi theta)^(phi)=cot theta],[Gamma_(theta theta)^(theta)=Gamma^(theta)_(theta phi)=0=Gamma^(phi)_(theta theta)=Gamma_(phi phi)^(phi).]:}\begin{align*} \Gamma_{\phi \phi}^{\theta} & =-\sin \theta \cos \theta \\ \Gamma_{\phi \theta}^{\phi} & =\cot \theta \tag{6}\\ \Gamma_{\theta \theta}^{\theta} & =\Gamma^{\theta}{ }_{\theta \phi}=0=\Gamma^{\phi}{ }_{\theta \theta}=\Gamma_{\phi \phi}^{\phi} . \end{align*}Γϕϕθ=sinθcosθ(6)Γϕθϕ=cotθΓθθθ=Γθθϕ=0=Γϕθθ=Γϕϕϕ.
Choose a suitable curvature component (one that is not automatically zero by reason of the elementary symmetry R μ ν α β = R [ μ ν ] [ α β ] R μ ν α β = R [ μ ν ] [ α β ] R_(mu nu alpha beta)=R_([mu nu][alpha beta])R_{\mu \nu \alpha \beta}=R_{[\mu \nu][\alpha \beta]}Rμναβ=R[μν][αβ], nor previously computed in another form, as by R μ ν α β = R α β μ ν R μ ν α β = R α β μ ν R_(mu nu alpha beta)=R_(alpha beta mu nu)R_{\mu \nu \alpha \beta}=R_{\alpha \beta \mu \nu}Rμναβ=Rαβμν ). In this two-dimensional example, there is only one choice (compared to 21 such computations in four dimensions); it is
R ϕ θ ϕ θ = Γ θ ϕ ϕ θ Γ ϕ θ θ θ ϕ + Γ θ k θ Γ ϕ ϕ k Γ k ϕ θ Γ ϕ θ k = Γ θ ϕ ϕ θ 0 + 0 Γ ϕ ϕ θ Γ ϕ θ ϕ = sin 2 θ cos 2 θ + sin θ cos θ cot θ R ϕ θ ϕ θ = Γ θ ϕ ϕ θ Γ ϕ θ θ θ ϕ + Γ θ k θ Γ ϕ ϕ k Γ k ϕ θ Γ ϕ θ k = Γ θ ϕ ϕ θ 0 + 0 Γ ϕ ϕ θ Γ ϕ θ ϕ = sin 2 θ cos 2 θ + sin θ cos θ cot θ {:[R_(phi theta phi)^(theta)=(delGamma^(theta)_(phi phi))/(del theta)-(delGamma_(phi theta theta)^(theta))/(del phi)+Gamma^(theta)_(k theta)Gamma_(phi phi)^(k)-Gamma_(k phi)^(theta)Gamma_(phi theta)^(k)],[=(delGamma^(theta)_(phi phi))/(del theta)-0+0-Gamma_(phi phi)^(theta)Gamma_(phi theta)^(phi)],[=sin^(2)theta-cos^(2)theta+sin theta cos theta cot theta]:}\begin{aligned} R_{\phi \theta \phi}^{\theta} & =\frac{\partial \Gamma^{\theta}{ }_{\phi \phi}}{\partial \theta}-\frac{\partial \Gamma_{\phi \theta \theta}^{\theta}}{\partial \phi}+\Gamma^{\theta}{ }_{k \theta} \Gamma_{\phi \phi}^{k}-\Gamma_{k \phi}^{\theta} \Gamma_{\phi \theta}^{k} \\ & =\frac{\partial \Gamma^{\theta}{ }_{\phi \phi}}{\partial \theta}-0+0-\Gamma_{\phi \phi}^{\theta} \Gamma_{\phi \theta}^{\phi} \\ & =\sin ^{2} \theta-\cos ^{2} \theta+\sin \theta \cos \theta \cot \theta \end{aligned}Rϕθϕθ=ΓθϕϕθΓϕθθθϕ+ΓθkθΓϕϕkΓkϕθΓϕθk=Γθϕϕθ0+0ΓϕϕθΓϕθϕ=sin2θcos2θ+sinθcosθcotθ
so
(7) R ϕ θ ϕ θ = sin 2 θ (7) R ϕ θ ϕ θ = sin 2 θ {:(7)R_(phi theta phi)^(theta)=sin^(2)theta:}\begin{equation*} R_{\phi \theta \phi}^{\theta}=\sin ^{2} \theta \tag{7} \end{equation*}(7)Rϕθϕθ=sin2θ
or
(8) R θ ϕ θ ϕ = 1 a 2 (8) R θ ϕ θ ϕ = 1 a 2 {:(8)R^(theta phi)_(theta phi)=(1)/(a^(2)):}\begin{equation*} R^{\theta \phi}{ }_{\theta \phi}=\frac{1}{a^{2}} \tag{8} \end{equation*}(8)Rθϕθϕ=1a2
Contraction gives the components of the Ricci tensor,
(9) R θ θ = R ϕ ϕ = 1 a 2 , R ϕ θ = 0 , (9) R θ θ = R ϕ ϕ = 1 a 2 , R ϕ θ = 0 , {:(9)R_(theta)^(theta)=R_(phi)^(phi)=(1)/(a^(2))","quadR_(phi)^(theta)=0",":}\begin{equation*} R_{\theta}^{\theta}=R_{\phi}^{\phi}=\frac{1}{a^{2}}, \quad R_{\phi}^{\theta}=0, \tag{9} \end{equation*}(9)Rθθ=Rϕϕ=1a2,Rϕθ=0,
and further contraction gives the curvature scalar
(10) R = 2 / a 2 . (10) R = 2 / a 2 . {:(10)R=2//a^(2).:}\begin{equation*} R=2 / a^{2} . \tag{10} \end{equation*}(10)R=2/a2.
A convenient orthonormal frame in this manifold is
(11) ω θ ^ = a d θ , ω ϕ ^ = a sin θ d ϕ . (11) ω θ ^ = a d θ , ω ϕ ^ = a sin θ d ϕ . {:(11)omega^( hat(theta))=ad theta","quadomega^( hat(phi))=a sin theta d phi.:}\begin{equation*} \boldsymbol{\omega}^{\hat{\theta}}=a \boldsymbol{d} \theta, \quad \boldsymbol{\omega}^{\hat{\phi}}=a \sin \theta \boldsymbol{d} \phi . \tag{11} \end{equation*}(11)ωθ^=adθ,ωϕ^=asinθdϕ.
More generally one writes ω α ^ = L α ^ β d x β ω α ^ = L α ^ β d x β omega^( hat(alpha))=L^( hat(alpha))_(beta)dx^(beta)\boldsymbol{\omega}^{\hat{\alpha}}=L^{\hat{\alpha}}{ }_{\beta} \boldsymbol{d} x^{\beta}ωα^=Lα^βdxβ. To transform the curvature tensor to orthonormal components in this simple but illuminating example of a diagonal metric requires a single normalization factor for each index on a tensor. Thus v θ ^ = a v θ , v ϕ ^ = a sin θ v ϕ , v θ ^ = a 1 v θ , v ϕ ^ = ( a sin θ ) 1 v ϕ v θ ^ = a v θ , v ϕ ^ = a sin θ v ϕ , v θ ^ = a 1 v θ , v ϕ ^ = ( a sin θ ) 1 v ϕ v^( hat(theta))=av^(theta),v^( hat(phi))=a sin thetav^(phi),v_( hat(theta))=a^(-1)v_(theta),v_( hat(phi))=(a sin theta)^(-1)v_(phi)v^{\hat{\theta}}=a v^{\theta}, v^{\hat{\phi}}=a \sin \theta v^{\phi}, v_{\hat{\theta}}=a^{-1} v_{\theta}, v_{\hat{\phi}}=(a \sin \theta)^{-1} v_{\phi}vθ^=avθ,vϕ^=asinθvϕ,vθ^=a1vθ,vϕ^=(asinθ)1vϕ. Similarly, from R θ ϕ θ ϕ = R θ ϕ θ ϕ = R^(theta)_(phi theta phi)=R^{\theta}{ }_{\phi \theta \phi}=Rθϕθϕ= sin 2 θ sin 2 θ sin^(2)theta\sin ^{2} \thetasin2θ one finds the components of the curvature tensor,
(12) R θ ^ ϕ ^ θ ^ ϕ ^ = 1 a 2 = R θ ^ ϕ ^ ϕ ^ ϕ ^ , (12) R θ ^ ϕ ^ θ ^ ϕ ^ = 1 a 2 = R θ ^ ϕ ^ ϕ ^ ϕ ^ , {:(12)R^( hat(theta)) hat(phi) hat(theta) hat(phi)=(1)/(a^(2))=R^( hat(theta) hat(phi) hat(phi)_( hat(phi)))",":}\begin{equation*} R^{\hat{\theta}} \hat{\phi} \hat{\theta} \hat{\phi}=\frac{1}{a^{2}}=R^{\hat{\theta} \hat{\phi} \hat{\phi}_{\hat{\phi}}}, \tag{12} \end{equation*}(12)Rθ^ϕ^θ^ϕ^=1a2=Rθ^ϕ^ϕ^ϕ^,
in the orthonormal frame.

Box 14.3 ANALYTICAL CALCULATIONS ON A COMPUTER

Research in gravitation physics and general relativity is sometimes beset by long calculations, requiring meticulous care, of such quantities as the Einstein and Riemann curvature tensors for a given metric, or the divergence of a given stressenergy tensor, or the Newman-Penrose tetrad equations under given algebraic assumptions. Such calculations are sufficiently straightforward and deductive in logical structure that they can be handled by a computer. Since 1966, computers have been generally taking over such tasks.
There are several computer languages in which the investigator can program his analytic calculations. The computer expert may wish to work in a machine-oriented language such as LISP [see, e.g., the work of Fletcher (1966) and of Hearn (1970)]. However, most appliers of relativity will prefer user-oriented languages such as REDUCE [created by Hearn (1970) and available for the IBM 360 and 370 , and the PDP 10 computers], ALAM [created by D'Inverno (1969) and available on Atlas computers], CAMAL [created by Barton, Bourne, and Fitch (1970) and available on Atlas computers], and FORMAC [created by Tobey et al. (1967) and available on IBM 7090, 7094, 360, and 370]. For a review of activity in this area, see Barton and Fitch (1971). Here we describe only FORMAC. It is the most widely available and widely used of the languages; but it is probably not the most powerful [see, e.g., D'Inverno (1969)]. FORMAC is to analytic work what the earliest and most primitive versions of FORTRAN were to numerical work.
FORMAC manipulates algebraic expressions involving: numerical constants, such as 1 / 3 1 / 3 1//31 / 31/3; symbolic constants, such as x x xxx or u u uuu; specific elementary functions, such as sin ( u ) sin ( u ) sin(u)\sin (u)sin(u) or exp ( x ) exp ( x ) exp(x)\exp (x)exp(x); and symbolic functions of several variables, such as f ( x , u ) f ( x , u ) f(x,u)f(x, u)f(x,u) or g ( u ) g ( u ) g(u)g(u)g(u). For example, it can add a x + b x 2 a x + b x 2 ax+bx^(2)a x+b x^{2}ax+bx2 to 2 x + 2 x + 2x+2 x+2x+ ( 3 + b ) x 2 ( 3 + b ) x 2 (3+b)x^(2)(3+b) x^{2}(3+b)x2 and get ( a + 2 ) x + ( 3 + 2 b ) x 2 ( a + 2 ) x + ( 3 + 2 b ) x 2 (a+2)x+(3+2b)x^(2)(a+2) x+(3+2 b) x^{2}(a+2)x+(3+2b)x2; it can take the partial derivative of x 2 u f ( x , u ) + cos ( x ) x 2 u f ( x , u ) + cos ( x ) x^(2)uf(x,u)+cos(x)x^{2} u f(x, u)+\cos (x)x2uf(x,u)+cos(x) with respect to x x xxx and get
2 x u f ( x , u ) + x 2 u f ( x , u ) / x sin ( x ) . 2 x u f ( x , u ) + x 2 u f ( x , u ) / x sin ( x ) . 2xuf(x,u)+x^(2)u del f(x,u)//del x-sin(x).2 x u f(x, u)+x^{2} u \partial f(x, u) / \partial x-\sin (x) .2xuf(x,u)+x2uf(x,u)/xsin(x).
It can do any algebraic or differential-calculus
computation that a human can do-but without making mistakes! Unfortunately, it cannot integrate analytically; integration requires inductive logic rather than deductive logic.
PL / 1 PL / 1 PL//1\mathrm{PL} / 1PL/1 is a language that can be used simultaneously with FORMAC or independently of it. PL / 1 PL / 1 PL//1\mathrm{PL} / 1PL/1 manipulates strings of characters-e.g., " Z / 1 × 29 + / Z / 1 × 29 + / Z//1xx29-+//\mathrm{Z} / 1 \times 29-+/Z/1×29+/. . It knows symbolic logic; it can tell whether two strings are identical; it can insert new characters into a string or remove old ones; but it does not know the rules of algebra or differential calculus. Thus, its primary use is as an adjunct to FORMAC (though from the viewpoint of the computer system FORMAC is an adjunct of PL/1).
FORMAC programs for evaluating Einstein's tensor in terms of given metric components and for doing other calculations are available from many past users [see, e.g., Fletcher, Clemens, Matzner, Thorne, and Zimmerman (1967); Ernst (1968); Harrison (1970)]. However, programming in FORMAC is sufficiently simple that one ordinarily does not have difficulty creating one's own program to do a given task. If a difficulty does arise, it may be because the analytic computation exhausts the core of the computer. It is easy to create an expression too large to fit in the core of any existing computer by several differentiations of an expression half a page long!
Users of FORMAC, confronted by coreexhaustion, have devised several ways to solve their problems. One is to remove unneeded parts of the program and of the FORMAC system from the core. Routines called PURGE and KILL have been developed for this purpose by Clemens and Matzner (1967). Another is to create the answer to a given calculation in manageable-sized pieces and output those pieces from the computer's core onto its disk. One must then add all the pieces together-a task that is impossible using FORMAC alone, or even FORMAC plus PL/1, but a task that James Hartle has solved [see Hartle and Thorne (1974)] by using a combination of FORMAC, PL/1, and IBM data-manipulation routines called SORT.
R μ ν α β R μ ν α β R_(mu nu alpha beta)R_{\mu \nu \alpha \beta}Rμναβ as a linear function of R R RRR, constructed from R R RRR and the metric alone, and with the correct contracted value R μ ν μ ν = R R μ ν μ ν = R R^(mu nu)_(mu nu)=RR^{\mu \nu}{ }_{\mu \nu}=RRμνμν=R. Establish a corresponding three-dimensional identity expressing R i j k R i j k R_(ijkℓ)R_{i j k \ell}Rijk in terms of the Ricci tensor R j k R j k R_(jk)R_{j k}Rjk and the metric.

Exercise 14.3. CURVATURE OF 3-SPHERE IN ORTHONORMAL FRAME

Compute the curvature tensor for a 3 -sphere
(14.2) d s 2 = a 2 [ d χ 2 + sin 2 χ ( d θ 2 + sin 2 θ d ϕ 2 ) ] (14.2) d s 2 = a 2 d χ 2 + sin 2 χ d θ 2 + sin 2 θ d ϕ 2 {:(14.2)ds^(2)=a^(2)[dchi^(2)+sin^(2)chi(dtheta^(2)+sin^(2)theta dphi^(2))]:}\begin{equation*} d s^{2}=a^{2}\left[d \chi^{2}+\sin ^{2} \chi\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right] \tag{14.2} \end{equation*}(14.2)ds2=a2[dχ2+sin2χ(dθ2+sin2θdϕ2)]
or for a 3-hyperboloid
(14.3) d s 2 = a 2 [ d χ 2 + sinh 2 χ ( d θ 2 + sin 2 θ d ϕ 2 ) ] (14.3) d s 2 = a 2 d χ 2 + sinh 2 χ d θ 2 + sin 2 θ d ϕ 2 {:(14.3)ds^(2)=a^(2)[dchi^(2)+sinh^(2)chi(dtheta^(2)+sin^(2)theta dphi^(2))]:}\begin{equation*} d s^{2}=a^{2}\left[d \chi^{2}+\sinh ^{2} \chi\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right] \tag{14.3} \end{equation*}(14.3)ds2=a2[dχ2+sinh2χ(dθ2+sin2θdϕ2)]
Convert the coordinate-based components R i j k i R i j k i R^(i)_(jki)R^{i}{ }_{j k i}Rijki to a corresponding orthonormal basis, R i j k R i j k R^(i_(jkℓ))R^{i_{j k \ell}}Rijk. Display R i j ^ k ^ ^ = R [ j ^ ] [ k ^ l ^ ] R i j ^ k ^ ^ = R [ j ^ ] [ k ^ l ^ ] R^( hat(ij) hat(k) hat(ℓ))=R^([ hat(j)])_([ hat(k) hat(l)])R^{\hat{i j} \hat{k} \hat{\ell}}=R^{[\hat{j}]}{ }_{[\hat{k} \hat{l}]}Rij^k^^=R[j^][k^l^] as a 3 × 3 3 × 3 3xx33 \times 33×3 matrix with appropriately labeled rows and columns.

§14.2. FORMING THE EINSTEIN TENSOR

The distribution of matter in space does not immediately tell all details of the local curvature of space, according to Einstein. The stress-energy tensor provides information only about a certain combination of components of the Riemann curvature tensor, the combination that makes up the Einstein tensor. Chapter 13 described two equivalent ways to calculate the Einstein tensor: (1) by successive contractions of the Riemann tensor
(14.4) R μ ν = R α μ α ν , R = g μ ν R μ ν , G μ ν = R μ ν 1 2 g μ ν R (14.4) R μ ν = R α μ α ν , R = g μ ν R μ ν , G μ ν = R μ ν 1 2 g μ ν R {:[(14.4)R_(mu nu)=R^(alpha)_(mu alpha nu)","quad R=g^(mu nu)R_(mu nu)","],[G_(mu nu)=R_(mu nu)-(1)/(2)g_(mu nu)R]:}\begin{gather*} R_{\mu \nu}=R^{\alpha}{ }_{\mu \alpha \nu}, \quad R=g^{\mu \nu} R_{\mu \nu}, \tag{14.4}\\ G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R \end{gather*}(14.4)Rμν=Rαμαν,R=gμνRμν,Gμν=Rμν12gμνR
[equations (13.48) and (13.49)]; (2) by forming the dual of the Riemann tensor and then contracting:
(14.5a) G α β γ δ ( R ) α β γ δ = ϵ α β μ ν R | μ ν | | ρ σ | ϵ ρ σ γ δ = δ ρ σ γ δ α β μ ν R | μ ν | | ρ σ | , (14.5b) G β δ = G α β α δ (14.5a) G α β γ δ R α β γ δ = ϵ α β μ ν R | μ ν | | ρ σ | ϵ ρ σ γ δ = δ ρ σ γ δ α β μ ν R | μ ν | | ρ σ | , (14.5b) G β δ = G α β α δ {:[(14.5a)G_(alpha beta)^(gamma delta)-=(^(**)R^(**))_(alpha beta)^(gamma delta)=epsilon_(alpha beta mu nu)R^(|mu nu|)_(|rho sigma|)epsilon^(rho sigma gamma delta)],[=-delta^(rho sigma gamma delta)_(alpha beta mu nu)R^(|mu nu|)_(|rho sigma|)","],[(14.5b)G_(beta)^(delta)=G_(alpha beta)^(alpha delta)]:}\begin{align*} G_{\alpha \beta}{ }^{\gamma \delta} \equiv\left({ }^{*} R^{*}\right)_{\alpha \beta}{ }^{\gamma \delta} & =\epsilon_{\alpha \beta \mu \nu} R^{|\mu \nu|}{ }_{|\rho \sigma|} \epsilon^{\rho \sigma \gamma \delta} \tag{14.5a}\\ & =-\delta^{\rho \sigma \gamma \delta}{ }_{\alpha \beta \mu \nu} R^{|\mu \nu|}{ }_{|\rho \sigma|}, \\ G_{\beta}{ }^{\delta} & =G_{\alpha \beta}{ }^{\alpha \delta} \tag{14.5b} \end{align*}(14.5a)Gαβγδ(R)αβγδ=ϵαβμνR|μν||ρσ|ϵρσγδ=δρσγδαβμνR|μν||ρσ|,(14.5b)Gβδ=Gαβαδ
[equations (13.46) and (13.47)]. A third method, usually superior to either of these, is discovered by combining equations ( 14.5 a , b 14.5 a , b 14.5a,b14.5 \mathrm{a}, \mathrm{b}14.5a,b ):
(14.6) G β δ = G δ β = δ δ ρ σ β μ ν R | μ ν | | ρ σ | . . (14.6) G β δ = G δ β = δ δ ρ σ β μ ν R | μ ν | | ρ σ | . . {:(14.6)G_(beta)^(delta)=G^(delta)_(beta)=-delta^(delta rho sigma)_(beta mu nu)R^(|mu nu|_(|rho sigma|^(.)).):}\begin{equation*} G_{\beta}{ }^{\delta}=G^{\delta}{ }_{\beta}=-\delta^{\delta \rho \sigma}{ }_{\beta \mu \nu} R^{|\mu \nu|_{|\rho \sigma|^{.}} .} \tag{14.6} \end{equation*}(14.6)Gβδ=Gδβ=δδρσβμνR|μν||ρσ|..
[Note: in any frame, orthonormal or not, the permutation tensor δ δ ρ σ β μ ν δ δ ρ σ β μ ν delta^(delta rho sigma)_(beta mu nu)\delta^{\delta \rho \sigma}{ }_{\beta \mu \nu}δδρσβμν has components
δ δ ρ σ β μ ν = δ β μ ν δ ρ σ = { + 1 if δ ρ σ is an even permutation of β μ ν , 1 if δ ρ σ is an odd permutation of β μ ν , 0 otherwise; δ δ ρ σ β μ ν = δ β μ ν δ ρ σ = + 1  if  δ ρ σ  is an even permutation of  β μ ν , 1  if  δ ρ σ  is an odd permutation of  β μ ν , 0  otherwise;  delta^(delta rho sigma)_(beta mu nu)=delta_(beta mu nu)^(delta rho sigma)={[+1" if "delta rho sigma" is an even permutation of "beta mu nu","],[-1" if "delta rho sigma" is an odd permutation of "beta mu nu","],[0" otherwise; "]:}\delta^{\delta \rho \sigma}{ }_{\beta \mu \nu}=\delta_{\beta \mu \nu}{ }^{\delta \rho \sigma}=\left\{\begin{array}{l} +1 \text { if } \delta \rho \sigma \text { is an even permutation of } \beta \mu \nu, \\ -1 \text { if } \delta \rho \sigma \text { is an odd permutation of } \beta \mu \nu, \\ 0 \text { otherwise; } \end{array}\right.δδρσβμν=δβμνδρσ={+1 if δρσ is an even permutation of βμν,1 if δρσ is an odd permutation of βμν,0 otherwise; 
Three ways to compute the Einstein tensor from the Riemann tensor
to see this, simply evaluate δ δ ρ σ β μ ν δ δ ρ σ β μ ν delta^(delta rho sigma)_(beta mu nu)\delta^{\delta \rho \sigma}{ }_{\beta \mu \nu}δδρσβμν using definition (3.50h) and using the components (8.10) of ϵ α β μ ν ϵ α β μ ν epsilon_(alpha beta mu nu)\epsilon_{\alpha \beta \mu \nu}ϵαβμν and ϵ ρ σ γ δ ϵ ρ σ γ δ epsilon^(rho sigma gamma delta)\epsilon^{\rho \sigma \gamma \delta}ϵρσγδ.] Equation (14.6) for the Einstein tensor, written out explicitly, reads
G 0 0 = ( R 12 12 + R 23 23 + R 31 31 ) , G 1 1 = ( R 02 02 + R 03 03 + R 23 23 , (14.7) G 0 1 = R 02 12 + R 03 13 , G 1 2 = R 10 20 + R 13 23 , G 0 0 = R 12 12 + R 23 23 + R 31 31 , G 1 1 = R 02 02 + R 03 03 + R 23 23 , (14.7) G 0 1 = R 02 12 + R 03 13 , G 1 2 = R 10 20 + R 13 23 , {:[G^(0)_(0)=-(R^(12)_(12)+R^(23)_(23)+R^(31)_(31))","],[G^(1)_(1)=-(R^(02)_(02)+R^(03)_(03)+R^(23)_(23),:}],[(14.7)G^(0)_(1)=R^(02)_(12)+R^(03)_(13)","],[G^(1)_(2)=R^(10)_(20)+R^(13)_(23)","]:}\begin{align*} & G^{0}{ }_{0}=-\left(R^{12}{ }_{12}+R^{23}{ }_{23}+R^{31}{ }_{31}\right), \\ & G^{1}{ }_{1}=-\left(R^{02}{ }_{02}+R^{03}{ }_{03}+R^{23}{ }_{23},\right. \\ & G^{0}{ }_{1}=R^{02}{ }_{12}+R^{03}{ }_{13}, \tag{14.7}\\ & G^{1}{ }_{2}=R^{10}{ }_{20}+R^{13}{ }_{23}, \end{align*}G00=(R1212+R2323+R3131),G11=(R0202+R0303+R2323,(14.7)G01=R0212+R0313,G12=R1020+R1323,
Standard method of computing curvature is wasteful
Ways to avoid "waste":
(1) geodesic Lagrangian method
(2) method of curvature 2-forms

§14.3. MORE EFFICIENT COMPUTATION

If the answer to a problem or the result of a computation is not simple, then there is no simple way to obtain it. But when a long computation gives a short answer, then one looks for a better method. Many of the best-known applications of general relativity present one with metric forms in which many of the components g μ ν , Γ μ α β g μ ν , Γ μ α β g_(mu nu),Gamma^(mu)_(alpha beta)g_{\mu \nu}, \Gamma^{\mu}{ }_{\alpha \beta}gμν,Γμαβ, and R μ v α β R μ v α β R^(mu)_(v alpha beta)R^{\mu}{ }_{v \alpha \beta}Rμvαβ are zero; for them the standard computation of the curvature (Box 14.2) involves much "wasted" effort. One computes many Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ that turn out to be zero. One checks off many terms in a sum like Γ μ ρ β Γ ρ α μ Γ μ ρ β Γ ρ α μ -Gamma^(mu)_(rho beta)Gamma^(rho)_(alpha mu)-\Gamma^{\mu}{ }_{\rho \beta} \Gamma^{\rho}{ }_{\alpha \mu}ΓμρβΓραμ that are zero, or cancel with others to give zero. Two alternative procedures are available to eliminate some of this "waste." The "geodesic Lagrangian" method provides an economical way to tabulate the Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ. The method of "curvature 2-forms" reorganizes the description from beginning to end, and computes both the connection and the curvature.
The geodesic Lagrangian method is only a moderate improvement over the standard method, but it also demands only a modest investment in the calculus of variations, an investment that pays off in any case in other contexts in the world of mathematics and physics. In contrast, the method of curvature 2 -forms is efficient, but demands a heavier investment in the mathematics of 1 -forms and 2 -forms than anyone would normally find needful for any introductory survey of relativity. Anyone facing several days' work at computing curvatures, however, would do well to learn the algorithm of the curvature 2 -forms.

§14.4. THE GEODESIC LAGRANGIAN METHOD

One normally thinks that the connection coefficients Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ must be known before one can write the geodesic equation
(14.8) x ¨ μ + Γ μ α β x ˙ α x ˙ β = 0 . (14.8) x ¨ μ + Γ μ α β x ˙ α x ˙ β = 0 . {:(14.8)x^(¨)^(mu)+Gamma^(mu)_(alpha beta)x^(˙)^(alpha)x^(˙)^(beta)=0.:}\begin{equation*} \ddot{x}^{\mu}+\Gamma^{\mu}{ }_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta}=0 . \tag{14.8} \end{equation*}(14.8)x¨μ+Γμαβx˙αx˙β=0.
(Here and below dots denote derivative with respect to the affine parameter, λ λ lambda\lambdaλ.) However, the argument can be reversed. Once the geodesic equations have been
written down, the connection coefficients can be read out of them. For instance, on the 2 -sphere as treated in Box 14.2, the geodesic equations are
(14.9) θ ¨ sin θ cos θ ϕ ˙ 2 = 0 ( 14.9 ϕ ) ϕ ¨ + 2 cot θ ϕ ˙ θ ˙ = 0 . (14.9) θ ¨ sin θ cos θ ϕ ˙ 2 = 0 ( 14.9 ϕ ) ϕ ¨ + 2 cot θ ϕ ˙ θ ˙ = 0 . {:[(14.9)theta^(¨)-sin theta cos thetaphi^(˙)^(2)=0],[((14.9 phi)")"phi^(¨)+2cot thetaphi^(˙)theta^(˙)=0.]:}\begin{align*} \ddot{\theta}-\sin \theta \cos \theta \dot{\phi}^{2} & =0 \tag{14.9}\\ \ddot{\phi}+2 \cot \theta \dot{\phi} \dot{\theta} & =0 . \tag{$14.9\phi$} \end{align*}(14.9)θ¨sinθcosθϕ˙2=0(14.9ϕ)ϕ¨+2cotθϕ˙θ˙=0.
The first equation here shows that Γ θ ϕ ϕ = sin θ cos θ Γ θ ϕ ϕ = sin θ cos θ Gamma^(theta)_(phi phi)=-sin theta cos theta\Gamma^{\theta}{ }_{\phi \phi}=-\sin \theta \cos \thetaΓθϕϕ=sinθcosθ; the second equation shows that Γ ϕ ϕ θ = Γ ϕ θ ϕ = cot θ Γ ϕ ϕ θ = Γ ϕ θ ϕ = cot θ Gamma^(phi)_(phi theta)=Gamma^(phi)_(theta phi)=cot theta\Gamma^{\phi}{ }_{\phi \theta}=\Gamma^{\phi}{ }_{\theta \phi}=\cot \thetaΓϕϕθ=Γϕθϕ=cotθ; and the absence of any further terms shows that all other Γ i j k Γ i j k Gamma^(i)_(jk)\Gamma^{i}{ }_{j k}Γijk are zero.
The first essential principle is thus clear: an explicit writing out of the geodesic equation is equivalent to a tabulation of all the connection coefficients Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ.
The second principle says more: one can write out the geodesic equation without ever having computed the Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ. In order to arrive at the equations for a geodesic (see Box 13.3), one need only recall that a geodesic is a parametrized curve that extremizes the integral
(14.10) I = 1 2 g μ ν x ˙ μ x ˙ ν d λ (14.10) I = 1 2 g μ ν x ˙ μ x ˙ ν d λ {:(14.10)I=(1)/(2)intg_(mu nu)x^(˙)^(mu)x^(˙)^(nu)d lambda:}\begin{equation*} I=\frac{1}{2} \int g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu} d \lambda \tag{14.10} \end{equation*}(14.10)I=12gμνx˙μx˙νdλ
in the sense
δ I = 0 δ I = 0 delta I=0\delta I=0δI=0
In practical applications of this variational principle, the first step is to rewrite equation (14.10) in the simplest possible form, inserting the specific values of g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν for the problem at hand. If one's interest attaches to the geodesics themselves, one can recognize many constants of motion even without carrying out any variations (see Chapter 25 on geodesic motion in Schwarzschild geometry, especially $ 25.2 $ 25.2 $25.2\$ 25.2$25.2 on conservation laws and constants of motion). For the purpose of computing the Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ, one proceeds to vary each coordinate in turn, obtaining four equations. Next these equations are rearranged so that their leading terms are x ¨ μ x ¨ μ x^(¨)^(mu)\ddot{x}^{\mu}x¨μ. In this form they must be precisely the geodesic equations (14.8). Consequently, the Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ are immediately available as the coefficients in these four equations. For the final step in computing curvature by this method, one returns to the standard method and to formulas of the type R Γ + Γ Γ R Γ + Γ Γ R∼del Gamma+Gamma GammaR \sim \partial \Gamma+\Gamma \GammaRΓ+ΓΓ, treated in the standard way (Box 14.2); and as the need arises for each Γ Γ Gamma\GammaΓ in turn, one scans the geodesic equation to find it. The procedure is best understood by following an example: Box 14.4 provides one.

Exercise 14.4. EINSTEIN EQUATIONS FOR THE CLOSED FRIEDMANN UNIVERSE CALCULATED BY USING THE GEODESIC LAGRANGIAN METHOD

The line element of interest here is (see Chapter 27)
d s 2 = d t 2 + a 2 ( t ) [ d χ 2 + sin 2 χ ( d θ 2 + sin 2 θ d ϕ 2 ) ] . d s 2 = d t 2 + a 2 ( t ) d χ 2 + sin 2 χ d θ 2 + sin 2 θ d ϕ 2 . ds^(2)=-dt^(2)+a^(2)(t)[dchi^(2)+sin^(2)chi(dtheta^(2)+sin^(2)theta dphi^(2))].d s^{2}=-d t^{2}+a^{2}(t)\left[d \chi^{2}+\sin ^{2} \chi\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right] .ds2=dt2+a2(t)[dχ2+sin2χ(dθ2+sin2θdϕ2)].
Geodesic Lagrangian method in 4 steps:
(1) write / in simple form
(2) vary / to get geodesic equation
(3) read off Γ α β γ Γ α β γ Gamma^(alpha)_(beta gamma)\Gamma^{\alpha}{ }_{\beta \gamma}Γαβγ
(4) compute R α β γ δ R α β γ δ R^(alpha)_(beta gamma delta)R^{\alpha}{ }_{\beta \gamma \delta}Rαβγδ etc. by standard method

EXERCISE

Box 14.4 GEODESIC LAGRANGIAN METHOD SHORTENS SOME CURVATURE COMPUTATIONS

Aim: Compute the curvature for the line element
(1) d s 2 = L 2 ( e 2 β d x 2 + e 2 β d y 2 ) 2 d u d v (1) d s 2 = L 2 e 2 β d x 2 + e 2 β d y 2 2 d u d v {:(1)ds^(2)=L^(2)(e^(2beta)dx^(2)+e^(-2beta)dy^(2))-2dudv:}\begin{equation*} d s^{2}=L^{2}\left(e^{2 \beta} d x^{2}+e^{-2 \beta} d y^{2}\right)-2 d u d v \tag{1} \end{equation*}(1)ds2=L2(e2βdx2+e2βdy2)2dudv
where L L LLL and β β beta\betaβ are functions of u u uuu only. [This metric is discussed as an example of a gravitational wave in §§35.9-35.12.]
Method: Obtain the Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ from the geodesic equations as inferred from the variational principle (14.10), then compute R μ ν α β Γ + Γ 2 R μ ν α β Γ + Γ 2 R^(mu)_(nu alpha beta)∼del Gamma+Gamma^(2)R^{\mu}{ }_{\nu \alpha \beta} \sim \partial \Gamma+\Gamma^{2}RμναβΓ+Γ2 as in Box 14.2.
Step 1. State the variational integral. For the metric under consideration, equation (14.10) requires δ I = 0 δ I = 0 delta I=0\delta I=0δI=0 for
(2) I = [ 1 2 L 2 ( e 2 β x ˙ 2 + e 2 β y ˙ 2 ) u ˙ v ˙ ] d λ . (2) I = 1 2 L 2 e 2 β x ˙ 2 + e 2 β y ˙ 2 u ˙ v ˙ d λ . {:(2)I=int[(1)/(2)L^(2)(e^(2beta)x^(˙)^(2)+e^(-2beta)y^(˙)^(2))-(u^(˙))(v^(˙))]d lambda.:}\begin{equation*} I=\int\left[\frac{1}{2} L^{2}\left(e^{2 \beta} \dot{x}^{2}+e^{-2 \beta} \dot{y}^{2}\right)-\dot{u} \dot{v}\right] d \lambda . \tag{2} \end{equation*}(2)I=[12L2(e2βx˙2+e2βy˙2)u˙v˙]dλ.
A world line that extremizes this integral is a geodesic.
Step 2: Vary the coordinates of the world line, one at a time, in their dependence on λ λ lambda\lambdaλ. First vary x ( λ ) x ( λ ) x(lambda)x(\lambda)x(λ), keeping fixed the functions y ( λ ) , u ( λ ) y ( λ ) , u ( λ ) y(lambda),u(lambda)y(\lambda), u(\lambda)y(λ),u(λ), and v ( λ ) v ( λ ) v(lambda)v(\lambda)v(λ). Then
δ I = ( L 2 e 2 β x ˙ ) δ x ˙ d λ = ( L 2 e 2 β x ˙ ) δ x d λ δ I = L 2 e 2 β x ˙ δ x ˙ d λ = L 2 e 2 β x ˙ δ x d λ delta I=int(L^(2)e^(2beta)(x^(˙)))deltax^(˙)d lambda=-int(L^(2)e^(2beta)(x^(˙)))^(*)delta xd lambda\delta I=\int\left(L^{2} e^{2 \beta} \dot{x}\right) \delta \dot{x} d \lambda=-\int\left(L^{2} e^{2 \beta} \dot{x}\right)^{\cdot} \delta x d \lambdaδI=(L2e2βx˙)δx˙dλ=(L2e2βx˙)δxdλ
The requirement that δ I = 0 δ I = 0 delta I=0\delta I=0δI=0 for this variation (among others) gives
0 = ( L 2 e 2 β x ˙ ) = L 2 e 2 β x ¨ + x ˙ u ˙ u ( L 2 e 2 β ) . 0 = L 2 e 2 β x ˙ = L 2 e 2 β x ¨ + x ˙ u ˙ u L 2 e 2 β . 0=(L^(2)e^(2beta)(x^(˙)))^(*)=L^(2)e^(2beta)x^(¨)+x^(˙)u^(˙)(del)/(del u)(L^(2)e^(2beta)).0=\left(L^{2} e^{2 \beta} \dot{x}\right)^{\cdot}=L^{2} e^{2 \beta} \ddot{x}+\dot{x} \dot{u} \frac{\partial}{\partial u}\left(L^{2} e^{2 \beta}\right) .0=(L2e2βx˙)=L2e2βx¨+x˙u˙u(L2e2β).
Varying y , u , v y , u , v y,u,vy, u, vy,u,v, in the same way gives
0 = ( L 2 e 2 β y ˙ ) = L 2 e 2 β y ¨ + y ˙ u ˙ u ( L 2 e 2 β ) , 0 = v ¨ + 1 2 x ˙ 2 u ( L 2 e 2 β ) + 1 2 y ˙ 2 u ( L 2 e 2 β ) , 0 = u ¨ . 0 = L 2 e 2 β y ˙ = L 2 e 2 β y ¨ + y ˙ u ˙ u L 2 e 2 β , 0 = v ¨ + 1 2 x ˙ 2 u L 2 e 2 β + 1 2 y ˙ 2 u L 2 e 2 β , 0 = u ¨ . {:[0=(L^(2)e^(-2beta)(y^(˙)))=L^(2)e^(-2beta)y^(¨)+y^(˙)u^(˙)(del)/(del u)(L^(2)e^(-2beta))","],[0=v^(¨)+(1)/(2)x^(˙)^(2)(del)/(del u)(L^(2)e^(2beta))+(1)/(2)y^(˙)^(2)(del)/(del u)(L^(2)e^(-2beta))","],[0=u^(¨).]:}\begin{aligned} & 0=\left(L^{2} e^{-2 \beta} \dot{y}\right)=L^{2} e^{-2 \beta} \ddot{y}+\dot{y} \dot{u} \frac{\partial}{\partial u}\left(L^{2} e^{-2 \beta}\right), \\ & 0=\ddot{v}+\frac{1}{2} \dot{x}^{2} \frac{\partial}{\partial u}\left(L^{2} e^{2 \beta}\right)+\frac{1}{2} \dot{y}^{2} \frac{\partial}{\partial u}\left(L^{2} e^{-2 \beta}\right), \\ & 0=\ddot{u} . \end{aligned}0=(L2e2βy˙)=L2e2βy¨+y˙u˙u(L2e2β),0=v¨+12x˙2u(L2e2β)+12y˙2u(L2e2β),0=u¨.
Step 3: Rearrange to get x ¨ μ x ¨ μ x^(¨)^(mu)\ddot{x}^{\mu}x¨μ leading terms. If this step is not straightforward, this method will not save time, and the technique of either Box 14.2 or Box 14.5 will be more suitable. In the example here, one quickly writes, using a prime for / u / u del//del u\partial / \partial u/u,
(3x) 0 = x ¨ + 2 ( L 1 L + β ) x ˙ u ˙ , (3y) 0 = y ¨ + 2 ( L 1 L β ) y ˙ u ˙ , (3v) 0 = v ¨ + ( L 2 e 2 β ) ( L 1 L + β ) x ˙ 2 + ( L 2 e 2 β ) ( L 1 L β ) y ˙ 2 , (3u) 0 = u ¨ . (3x) 0 = x ¨ + 2 L 1 L + β x ˙ u ˙ , (3y) 0 = y ¨ + 2 L 1 L β y ˙ u ˙ , (3v) 0 = v ¨ + L 2 e 2 β L 1 L + β x ˙ 2 + L 2 e 2 β L 1 L β y ˙ 2 , (3u) 0 = u ¨ . {:[(3x)0=x^(¨)+2(L^(-1)L^(')+beta^('))x^(˙)u^(˙)","],[(3y)0=y^(¨)+2(L^(-1)L^(')-beta^('))y^(˙)u^(˙)","],[(3v)0=v^(¨)+(L^(2)e^(2beta))(L^(-1)L^(')+beta^('))x^(˙)^(2)+(L^(2)e^(-2beta))(L^(-1)L^(')-beta^('))y^(˙)^(2)","],[(3u)0=u^(¨).]:}\begin{align*} & 0=\ddot{x}+2\left(L^{-1} L^{\prime}+\beta^{\prime}\right) \dot{x} \dot{u}, \tag{3x}\\ & 0=\ddot{y}+2\left(L^{-1} L^{\prime}-\beta^{\prime}\right) \dot{y} \dot{u}, \tag{3y}\\ & 0=\ddot{v}+\left(L^{2} e^{2 \beta}\right)\left(L^{-1} L^{\prime}+\beta^{\prime}\right) \dot{x}^{2}+\left(L^{2} e^{-2 \beta}\right)\left(L^{-1} L^{\prime}-\beta^{\prime}\right) \dot{y}^{2}, \tag{3v}\\ & 0=\ddot{u} . \tag{3u} \end{align*}(3x)0=x¨+2(L1L+β)x˙u˙,(3y)0=y¨+2(L1Lβ)y˙u˙,(3v)0=v¨+(L2e2β)(L1L+β)x˙2+(L2e2β)(L1Lβ)y˙2,(3u)0=u¨.
Step 3': Interpret these equations as a tabulation of Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ. Equations (3) are the standard equations for a geodesic,
x ¨ μ + Γ α β μ x ˙ α x ˙ β = 0 x ¨ μ + Γ α β μ x ˙ α x ˙ β = 0 x^(¨)^(mu)+Gamma_(alpha beta)^(mu)x^(˙)^(alpha)x^(˙)^(beta)=0\ddot{x}^{\mu}+\Gamma_{\alpha \beta}^{\mu} \dot{x}^{\alpha} \dot{x}^{\beta}=0x¨μ+Γαβμx˙αx˙β=0
Therefore it is enough to scan them to find the value of any desired Γ Γ Gamma\GammaΓ. For instance Γ x y u Γ x y u Gamma^(x)_(yu)\Gamma^{x}{ }_{y u}Γxyu must appear in the coefficient ( Γ x y u + Γ x u y ) = 2 Γ x y u Γ x y u + Γ x u y = 2 Γ x y u (Gamma^(x)_(yu)+Gamma^(x)_(uy))=2Gamma^(x)_(yu)\left(\Gamma^{x}{ }_{y u}+\Gamma^{x}{ }_{u y}\right)=2 \Gamma^{x}{ }_{y u}(Γxyu+Γxuy)=2Γxyu of the y ˙ u ˙ y ˙ u ˙ y^(˙)u^(˙)\dot{y} \dot{u}y˙u˙ term in the equation for x ¨ x ¨ x^(¨)\ddot{x}x¨. But no y ˙ u ˙ y ˙ u ˙ y^(˙)u^(˙)\dot{y} \dot{u}y˙u˙ term appears in equation (3x). Therefore Γ x y u Γ x y u Gamma^(x)_(yu)\Gamma^{x}{ }_{y u}Γxyu is zero in this example. Note that equations (3) are simple, in the sense that they contain few terms; therefore most of the Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ must be zero. For instance, it follows from equation ( 3 u 3 u 3u3 u3u ) that all ten Γ u α β Γ u α β Gamma^(u)_(alpha beta)\Gamma^{u}{ }_{\alpha \beta}Γuαβ are zero. The only non-zero Γ Γ Gamma^(')\Gamma^{\prime}Γ 's are Γ x x u = Γ x u x = Γ x x u = Γ x u x = Gamma^(x)_(xu)=Gamma^(x)_(ux)=\Gamma^{x}{ }_{x u}=\Gamma^{x}{ }_{u x}=Γxxu=Γxux= ( L 1 L + β L 1 L + β L^(-1)L^(')+beta^(')L^{-1} L^{\prime}+\beta^{\prime}L1L+β ) from equation (3x), Γ y y u = Γ y u y = ( L 1 L β ) Γ y y u = Γ y u y = L 1 L β Gamma^(y)_(yu)=Gamma^(y)_(uy)=(L^(-1)L^(')-beta^('))\Gamma^{y}{ }_{y u}=\Gamma^{y}{ }_{u y}=\left(L^{-1} L^{\prime}-\beta^{\prime}\right)Γyyu=Γyuy=(L1Lβ) from equation (3y), and Γ v x x Γ v x x Gamma^(v)_(xx)\Gamma^{v}{ }_{x x}Γvxx and Γ v y y Γ v y y Gamma^(v)_(yy)\Gamma^{v}{ }_{y y}Γvyy from equation (3v).
Step 4: Compute each R μ v α β R μ v α β R^(mu)_(v alpha beta)R^{\mu}{ }_{v \alpha \beta}Rμvαβ, etc. There is little relief from routine in systematically applying equation (3) from Box 14.2. One must list 21 components R μ ν α β R μ ν α β R^(mu)_(nu alpha beta)R^{\mu}{ }_{\nu \alpha \beta}Rμναβ that are not related by any of the symmetries R μ ν α β = R α β μ ν = R μ ν β α R μ ν α β = R α β μ ν = R μ ν β α R_(mu nu alpha beta)=R_(alpha beta mu nu)=-R_(mu nu beta alpha)R_{\mu \nu \alpha \beta}=R_{\alpha \beta \mu \nu}=-R_{\mu \nu \beta \alpha}Rμναβ=Rαβμν=Rμνβα, and compute each. In the example here, one notes that Γ u α β = 0 Γ u α β = 0 Gamma^(u)_(alpha beta)=0\Gamma^{u}{ }_{\alpha \beta}=0Γuαβ=0 implies R u α β γ = R v α β γ = 0 R u α β γ = R v α β γ = 0 R^(u)_(alpha beta gamma)=-R_(v alpha beta gamma)=0R^{u}{ }_{\alpha \beta \gamma}=-R_{v \alpha \beta \gamma}=0Ruαβγ=Rvαβγ=0. Therefore 15 of the list of 21 vanish at one swat. The list then is:
\begin{align*} R_{v \alpha \beta \gamma} & =-R^{u}{ }_{\alpha \beta \gamma} \end{aligned}=0, \quad \begin{aligned} R_{u x u x} & =-R^{v}{ }_{x u x} \end{align*}=-\left(\Gamma^{v}{ }_{x x}\right)^{\prime}+\Gamma^{v}{ }_{x x} \Gamma^{x}{ }_{x u} .\begin{align*} ended with \end{aligned}
One can now calculate the Einstein tensor via equation (14.7). In the example here, however, it is equally simple to form first the Ricci tensor by the straightforward contraction R μ α μ β R μ α μ β R^(mu)_(alpha mu beta)R^{\mu}{ }_{\alpha \mu \beta}Rμαμβ. Only μ = x μ = x mu=x\mu=xμ=x and μ = y μ = y mu=y\mu=yμ=y give any contribution, because no superscript index can be a u u uuu, and no subscript a v v vvv. Thus one finds
R u u = 2 [ L 1 L + β 2 ] , (5) all other R α β = 0 , R u u = 2 L 1 L + β 2 , (5)  all other  R α β = 0 , {:[R_(uu)=-2[L^(-1)L^('')+beta^('2)]","],[(5)" all other "R_(alpha beta)=0","]:}\begin{gather*} R_{u u}=-2\left[L^{-1} L^{\prime \prime}+\beta^{\prime 2}\right], \\ \text { all other } R_{\alpha \beta}=0, \tag{5} \end{gather*}Ruu=2[L1L+β2],(5) all other Rαβ=0,
and
(6) R = 0 (6) R = 0 {:(6)R=0:}\begin{equation*} R=0 \tag{6} \end{equation*}(6)R=0
From this last result, it follows that here the desired Einstein tensor is identical with the Ricci tensor.
(a) Set up the variational integral (14.10) for a geodesic in this metric, then successively vary t , χ , θ t , χ , θ t,chi,thetat, \chi, \thetat,χ,θ, and ϕ ϕ phi\phiϕ to obtain, after some rearrangement, four equations 0 = t ¨ + , 0 = 0 = t ¨ + , 0 = 0=t^(¨)+cdots,0=0=\ddot{t}+\cdots, 0=0=t¨+,0= χ ¨ + χ ¨ + chi^(¨)+cdots\ddot{\chi}+\cdotsχ¨+, etc. displaying the Γ Γ Gamma\GammaΓ 's in the form of equation (14.8).
(b) Use this display as a table of Γ Γ Gamma^(')\Gamma^{\prime}Γ s to compute R t χ μ ν R t χ μ ν R^(t)_(chi mu nu)R^{t}{ }_{\chi \mu \nu}Rtχμν and R χ θ μ ν R χ θ μ ν R^(chi_(theta mu nu))R^{\chi_{\theta \mu \nu}}Rχθμν, of which only R t χ t χ R t χ t χ R^(t)_(chi t chi)R^{t}{ }_{\chi t \chi}Rtχtχ and R x θ χ θ R x θ χ θ R^(x)_(theta chi theta)R^{x}{ }_{\theta \chi \theta}Rxθχθ are non-zero (consequence of the complete equivalence of all directions tangent to the χ θ ϕ χ θ ϕ chi theta phi\chi \theta \phiχθϕ sphere).
(c) Convert to an orthonormal frame with ω i ^ = d t , ω x ^ = a d χ , ω θ ^ = ω i ^ = d t , ω x ^ = a d χ , ω θ ^ = omega^( hat(i))=dt,omega^( hat(x))=ad chi,omega^( hat(theta))=\boldsymbol{\omega}^{\hat{i}}=\boldsymbol{d} t, \boldsymbol{\omega}^{\hat{x}}=a \boldsymbol{d} \chi, \boldsymbol{\omega}^{\hat{\theta}}=ωi^=dt,ωx^=adχ,ωθ^= ?, ω ϕ ^ = ω ϕ ^ = omega^( hat(phi))=\boldsymbol{\omega}^{\hat{\phi}}=ωϕ^= ?, and list R i χ ^ t ^ χ ^ R i χ ^ t ^ χ ^ R^(i hat(chi)) hat(t) hat(chi)R^{i \hat{\chi}} \hat{t} \hat{\chi}Riχ^t^χ^ and R χ ^ θ ^ χ ^ θ ^ R χ ^ θ ^ χ ^ θ ^ R^( hat(chi)_( hat(theta)))_( hat(chi))^( hat(theta))R^{\hat{\chi}_{\hat{\theta}}}{ }_{\hat{\chi}}^{\hat{\theta}}Rχ^θ^χ^θ^. Explain why all other components are known by symmetry in terms of these two.
(d) Calculate, using equations (14.7), all independent components of the Einstein tensor G μ ^ γ ^ G μ ^ γ ^ G^( hat(mu)_( hat(gamma)))G^{\hat{\mu}_{\hat{\gamma}}}Gμ^γ^. [Answer: See Box 14.5.]

§14.5. CURVATURE 2-FORMS

In electrodynamics the abstract notation
F = d A F = d A F=dA\boldsymbol{F}=\boldsymbol{d} \boldsymbol{A}F=dA
saves space compared to the explicit notation
F 31 = A 1 x 3 A 3 x 1 F 12 = A 2 x 1 A 1 x 2 , etc. (six equations); F 31 = A 1 x 3 A 3 x 1 F 12 = A 2 x 1 A 1 x 2 ,  etc. (six equations);  {:[F_(31)=(delA_(1))/(delx^(3))-(delA_(3))/(delx^(1))],[F_(12)=(delA_(2))/(delx^(1))-(delA_(1))/(delx^(2))],[ dots","" etc. (six equations); "]:}\begin{aligned} & F_{31}=\frac{\partial A_{1}}{\partial x^{3}}-\frac{\partial A_{3}}{\partial x^{1}} \\ & F_{12}=\frac{\partial A_{2}}{\partial x^{1}}-\frac{\partial A_{1}}{\partial x^{2}} \\ & \ldots, \text { etc. (six equations); } \end{aligned}F31=A1x3A3x1F12=A2x1A1x2, etc. (six equations); 
Concepts needed for method of curvature 2 -forms
there is no reason to shun similar economies in dealing with the dynamics of geometry. Cartan introduced the decisive ideas, seen above, of differential forms (where a simple object replaces a listing of four components; thus, σ = σ μ d x μ σ = σ μ d x μ sigma=sigma_(mu)dx^(mu)\boldsymbol{\sigma}=\sigma_{\mu} \boldsymbol{d} x^{\mu}σ=σμdxμ ), and of the exterior derivative d d d\boldsymbol{d}d. He went on ( 1928 , 1946 ) ( 1928 , 1946 ) (1928,1946)(1928,1946)(1928,1946) to package the 21 components R μ ν α β R μ ν α β R_(mu nu alpha beta)R_{\mu \nu \alpha \beta}Rμναβ of the curvature tensor into six curvature 2 -forms,
R μ ν = R ν μ . R μ ν = R ν μ . R^(mu nu)=-R^(nu mu).\mathscr{R}^{\mu \nu}=-\mathscr{R}^{\nu \mu} .Rμν=Rνμ.
Regarded purely as notation, these 2 -forms automatically produce a profit. They cut down the weight of paper work required to list one's answer after one has it. They also provide a route into deeper insight on "curvature as a geometric object," although that is not the objective of immediate concern in this chapter.
Cartan's exterior derivative d d d\boldsymbol{d}d automatically effects many cancelations in the calculation of curvature. It often cancels terms before they ever need to be evaluated.
Extension of Cartan's calculus from electromagnetism and other applications (Chapter 4) to the analysis of curvature (this chapter) requires two minor additions to the armament of forms and exterior derivative: (1) the idea of a vector-valued (or tensor-valued) exterior differential form; and (2) a corresponding generalization
of the exterior derivative d d d\boldsymbol{d}d. This section uses both these tools in deriving the key formulas (14.18), (14.25), (14.31), and (14.32). Once derived, however, these formulas demand no more than the standard exterior derivative for all applications and for all calculations of curvature ( $ 14.6 $ 14.6 $14.6\$ 14.6$14.6 and Box 14.5).
The extended exterior derivative leads to nothing new in the first two contexts to which one applies it: a scalar function (" 0 -form") and a vector field ("vector-valued 0 -form"). Thus, take any function f f fff. Its derivative in an unspecified direction is a 1 -form; or, to make a new distinction that will soon become meaningful, a "scalarvalued 1 -form." Specify the direction in which differentiation is to occur ("fill in the slot in the 1 -form"). Thereby obtain the ordinary derivative as it applies to a function
(14.11) d f , u = u f (14.11) d f , u = u f {:(14.11)(:df","u:)=del_(u)f:}\begin{equation*} \langle\boldsymbol{d} f, \boldsymbol{u}\rangle=\partial_{\boldsymbol{u}} f \tag{14.11} \end{equation*}(14.11)df,u=uf
Next, take any vector field v v v\boldsymbol{v}v. Its covariant derivative in an unspecified direction is a "vector-valued 1 -form." Specify the direction u u u\boldsymbol{u}u in which differentiation is to occur ("fill in the slot in the 1 -form"). Thereby obtain the covariant derivative
(14.12a) d v , u u v (14.12a) d v , u u v {:(14.12a)(:dv","u:)-=grad_(u)v:}\begin{equation*} \langle\boldsymbol{d v}, u\rangle \equiv \nabla_{u} v \tag{14.12a} \end{equation*}(14.12a)dv,uuv
This object too is not new; it is the covariant derivative of the vector v v v\boldsymbol{v}v taken in the direction of the vector u u u\boldsymbol{u}u. When one abstracts away from any special choice of the direction of differentiation u u u\boldsymbol{u}u, one finds an expression that one has encountered before, though not under its new name of "vector-valued 1 -form." This expression measures the covariant derivative of the vector v v v\boldsymbol{v}v in an unspecified direction ("slot for direction not yet filled in"). From a look at (14.12a), one sees that this extended exterior derivative is applied to v v v\boldsymbol{v}v, without reference to u u u\boldsymbol{u}u, is
(14.12b) d v = v (14.12b) d v = v {:(14.12b)dv=grad v:}\begin{equation*} d v=\nabla v \tag{14.12b} \end{equation*}(14.12b)dv=v
Similarly, for any "tensor-valued 0-form" [i.e. ( n 0 ) ( n 0 ) ((n)/(0))\binom{n}{0}(n0) tensor] S , d S S S , d S S S,dS-=grad S\boldsymbol{S}, \boldsymbol{d} \boldsymbol{S} \equiv \boldsymbol{\nabla} \boldsymbol{S}S,dSS.
Before proceeding further with the exterior (soon to be marked as "antisymmetric") differentiation of tensors, write down a formula (see exercise 14.5) for the exterior (antisymmetric) derivative of a product of forms:
(14.13a) d ( α β ) = ( d α ) β + ( 1 ) p α d β (14.13a) d ( α β ) = ( d α ) β + ( 1 ) p α d β {:(14.13a)d(alpha^^beta)=(d alpha)^^beta+(-1)^(p)alpha^^d beta:}\begin{equation*} \boldsymbol{d}(\boldsymbol{\alpha} \wedge \boldsymbol{\beta})=(\boldsymbol{d} \alpha) \wedge \boldsymbol{\beta}+(-1)^{p} \boldsymbol{\alpha} \wedge \boldsymbol{d} \boldsymbol{\beta} \tag{14.13a} \end{equation*}(14.13a)d(αβ)=(dα)β+(1)pαdβ
where α α alpha\boldsymbol{\alpha}α is a p p ppp-form and β β beta\boldsymbol{\beta}β is a q q qqq-form.
Now extend the exterior derivative from elementary forms to the exterior product of a tensor-valued p p ppp-form S S S\boldsymbol{S}S with any ordinary q q qqq-form, β β beta\boldsymbol{\beta}β; thus,
(14.13b) d ( S β ) = d S β + ( 1 ) p S d β (14.13b) d ( S β ) = d S β + ( 1 ) p S d β {:(14.13b)d(S^^beta)=dS^^beta+(-1)^(p)S^^d beta:}\begin{equation*} \boldsymbol{d}(\boldsymbol{S} \wedge \boldsymbol{\beta})=\boldsymbol{d} \boldsymbol{S} \wedge \boldsymbol{\beta}+(-1)^{p} \boldsymbol{S} \wedge \boldsymbol{d} \boldsymbol{\beta} \tag{14.13b} \end{equation*}(14.13b)d(Sβ)=dSβ+(1)pSdβ
This equation can be regarded as a general definition of the extended exterior derivative. For example, if S S S\boldsymbol{S}S is a tensor-valued 2-form, S = S α β | γ δ | e α e β d x γ d x δ S = S α β | γ δ | e α e β d x γ d x δ S=S^(alpha beta)_(|gamma delta|)e_(alpha)e_(beta)dx^(gamma)^^dx^(delta)\boldsymbol{S}=S^{\alpha \beta}{ }_{|\gamma \delta|} \boldsymbol{e}_{\alpha} \boldsymbol{e}_{\beta} \boldsymbol{d} x^{\gamma} \wedge \boldsymbol{d} x^{\delta}S=Sαβ|γδ|eαeβdxγdxδ, then equation (14.13b) says
d S = d [ ( e α e β S α β | γ δ | ) ( d x γ d x δ ) ] = d ( e α e β S α β | γ δ | ) ( d x γ d x δ ) d S = d e α e β S α β | γ δ | d x γ d x δ = d e α e β S α β | γ δ | d x γ d x δ dS=d[(e_(alpha)e_(beta)S^(alpha beta)_(|gamma delta|))(dx^(gamma)^^dx^(delta))]=d(e_(alpha)e_(beta)S^(alpha beta)_(|gamma delta|))^^(dx^(gamma)^^dx^(delta))\boldsymbol{d} \boldsymbol{S}=\boldsymbol{d}\left[\left(\boldsymbol{e}_{\alpha} \boldsymbol{e}_{\beta} S^{\alpha \beta}{ }_{|\gamma \delta|}\right)\left(\boldsymbol{d} x^{\gamma} \wedge \boldsymbol{d} x^{\delta}\right)\right]=\boldsymbol{d}\left(\boldsymbol{e}_{\alpha} \boldsymbol{e}_{\beta} S^{\alpha \beta}{ }_{|\gamma \delta|}\right) \wedge\left(\boldsymbol{d} x^{\gamma} \wedge \boldsymbol{d} x^{\delta}\right)dS=d[(eαeβSαβ|γδ|)(dxγdxδ)]=d(eαeβSαβ|γδ|)(dxγdxδ)
Extended exterior derivative:
(1) acting on a scalar
(2) acting on a vector
As another example, use (14.13b) to calculate d ( u σ ) d ( u σ ) d(u sigma)\boldsymbol{d}(\boldsymbol{u} \boldsymbol{\sigma})d(uσ), where u u u\boldsymbol{u}u is a vector-valued 0 -form (vector) and σ σ sigma\boldsymbol{\sigma}σ is a scalar-valued 1 -form ( 1 -form):
d ( u σ ) = ( d u ) σ + u d σ . d ( u σ ) = ( d u ) σ + u d σ . d(u sigma)=(du)^^sigma+ud sigma.\boldsymbol{d}(\boldsymbol{u} \sigma)=(\boldsymbol{d} \boldsymbol{u}) \wedge \boldsymbol{\sigma}+\boldsymbol{u} \boldsymbol{d} \sigma .d(uσ)=(du)σ+udσ.
If one were following the practice of earlier chapters, one would have written u σ u σ u ox sigma\boldsymbol{u} \otimes \sigmauσ where u σ u σ u sigma\boldsymbol{u} \boldsymbol{\sigma}uσ appears here, u d σ u d σ u ox d sigma\boldsymbol{u} \otimes \boldsymbol{d} \sigmaudσ instead of u d σ u d σ ud sigma\boldsymbol{u} \boldsymbol{d} \boldsymbol{\sigma}udσ, and e α e β e α e β e_(alpha)oxe_(beta)\boldsymbol{e}_{\alpha} \otimes \boldsymbol{e}_{\beta}eαeβ instead of e α e β e α e β e_(alpha)e_(beta)\boldsymbol{e}_{\alpha} \boldsymbol{e}_{\beta}eαeβ. However, to avoid overcomplication in the notation, all such tensor product symbols are omitted here and hereafter.
Equations (14.12) and (14.13) do more than define the (extended) exterior derivative d d d\boldsymbol{d}d and provide a way to use it in computations. They also allow one to define and calculate the antisymmetrized second derivatives, e.g., d 2 v d 2 v d^(2)v\boldsymbol{d}^{2} \boldsymbol{v}d2v. The relation
d 2 v = R v d 2 v = R v d^(2)v=Rv\boldsymbol{d}^{2} \boldsymbol{v}=\mathscr{R} \boldsymbol{v}d2v=Rv
where v v v\boldsymbol{v}v is a vector will then introduce the "operator-valued" or "( 1 1 1 ) 1 1 1 {:1[1],[1])\left.1 \begin{array}{l}1 \\ 1\end{array}\right)111)-tensor valued" curvature 2 -form R R R\mathscr{R}R. The notation of the extended exterior derivative puts a new look on the old apparatus of base vectors and parallel transport, and opens a way to calculate the curvature 2 -form R R R\mathscr{R}R.
Let the vector field v v v\boldsymbol{v}v be expanded in terms of some field of basis vectors e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ; thus
v = e μ v μ v = e μ v μ v=e_(mu)v^(mu)\boldsymbol{v}=\boldsymbol{e}_{\mu} v^{\mu}v=eμvμ
Then the exterior derivative of this vector is
d v = d e μ v μ + e μ d v μ d v = d e μ v μ + e μ d v μ dv=de_(mu)v^(mu)+e_(mu)dv^(mu)\boldsymbol{d} \boldsymbol{v}=\boldsymbol{d} \boldsymbol{e}_{\mu} v^{\mu}+\boldsymbol{e}_{\mu} \boldsymbol{d} v^{\mu}dv=deμvμ+eμdvμ
Expand the typical vector-valued 1-form d e μ d e μ de_(mu)\boldsymbol{d e}_{\mu}deμ in the form
Definition of ω μ v ω μ v omega^(mu)_(v)\boldsymbol{\omega}^{\mu}{ }_{v}ωμv
(14.14) d e μ = e ν ω ν μ . (14.14) d e μ = e ν ω ν μ . {:(14.14)de_(mu)=e_(nu)omega^(nu)_(mu).:}\begin{equation*} \boldsymbol{d} \boldsymbol{e}_{\mu}=\boldsymbol{e}_{\nu} \boldsymbol{\omega}^{\nu}{ }_{\mu} . \tag{14.14} \end{equation*}(14.14)deμ=eνωνμ.
Here the "components" ω ν μ ω ν μ omega^(nu)_(mu)\boldsymbol{\omega}^{\nu}{ }_{\mu}ωνμ in the expansion of d e μ d e μ de_(mu)\boldsymbol{d} \boldsymbol{e}_{\mu}deμ are 1 -forms. Recall from equation (10.13) that the typical ω ν μ ω ν μ omega^(nu)_(mu)\boldsymbol{\omega}^{\nu}{ }_{\mu}ωνμ is related to the connection coefficients by
(14.15) ω v μ = Γ v μ λ ω λ . (14.15) ω v μ = Γ v μ λ ω λ . {:(14.15)omega^(v)_(mu)=Gamma^(v)_(mu lambda)omega^(lambda).:}\begin{equation*} \boldsymbol{\omega}^{v}{ }_{\mu}=\Gamma^{v}{ }_{\mu \lambda} \boldsymbol{\omega}^{\lambda} . \tag{14.15} \end{equation*}(14.15)ωvμ=Γvμλωλ.
Therefore the expansion of the "vector" (really, "vector-valued 1 -form") is
(14.16) d v = e μ ( d v μ + ω μ ν v ν ) (14.16) d v = e μ d v μ + ω μ ν v ν {:(14.16)dv=e_(mu)(dv^(mu)+omega^(mu)_(nu)v^(nu)):}\begin{equation*} \boldsymbol{d} \boldsymbol{v}=\boldsymbol{e}_{\mu}\left(\boldsymbol{d} v^{\mu}+\boldsymbol{\omega}^{\mu}{ }_{\nu} v^{\nu}\right) \tag{14.16} \end{equation*}(14.16)dv=eμ(dvμ+ωμνvν)
Now differentiate once again to find
d 2 v = d e α ( d v α + ω α ν v v ) + e μ ( d 2 v μ + d ω μ v ω μ ν d v v ) = e μ ( ω μ α d v α + ω μ α ω α 2 v v + d 2 v μ + d ω μ ν v v ω μ α d v α ) . d 2 v = d e α d v α + ω α ν v v + e μ d 2 v μ + d ω μ v ω μ ν d v v = e μ ω μ α d v α + ω μ α ω α 2 v v + d 2 v μ + d ω μ ν v v ω μ α d v α . {:[d^(2)v=de_(alpha)^^(dv^(alpha)+omega^(alpha)_(nu)v^(v))],[+e_(mu)(d^(2)v^(mu)+domega^(mu)^(')^(v)-omega^(mu)_(nu)^^dv^(v))],[=e_(mu)(omega^(mu)_(alpha)^^dv^(alpha)+omega^(mu)_(alpha)^^omega^(alpha)^(2)v^(v):}],[{:+d^(2)v^(mu)+domega^(mu)_(nu)v^(v)-omega^(mu)_(alpha)^^dv^(alpha)).]:}\begin{aligned} \boldsymbol{d}^{2} \boldsymbol{v}= & \boldsymbol{d} \boldsymbol{e}_{\alpha} \wedge\left(\boldsymbol{d} v^{\alpha}+\boldsymbol{\omega}^{\alpha}{ }_{\nu} v^{v}\right) \\ & +\boldsymbol{e}_{\mu}\left(\boldsymbol{d}^{2} v^{\mu}+\boldsymbol{d} \omega^{\mu}{ }^{\prime}{ }^{v}-\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{d} v^{v}\right) \\ = & \boldsymbol{e}_{\mu}\left(\boldsymbol{\omega}^{\mu}{ }_{\alpha} \wedge \boldsymbol{d} v^{\alpha}+\boldsymbol{\omega}^{\mu}{ }_{\alpha} \wedge \boldsymbol{\omega}^{\alpha}{ }^{2} v^{v}\right. \\ & \left.+\boldsymbol{d}^{2} v^{\mu}+\boldsymbol{d} \boldsymbol{\omega}^{\mu}{ }_{\nu} v^{v}-\boldsymbol{\omega}^{\mu}{ }_{\alpha} \wedge \boldsymbol{d} v^{\alpha}\right) . \end{aligned}d2v=deα(dvα+ωανvv)+eμ(d2vμ+dωμvωμνdvv)=eμ(ωμαdvα+ωμαωα2vv+d2vμ+dωμνvvωμαdvα).
The simplifications made here use (1) the equation (14.14), for a second time; and (2) the product rule (14.13a), which introduced the minus sign in the last term, ready
to cancel the first term. Now consider the term d 2 v μ d 2 v μ d^(2)v^(mu)\boldsymbol{d}^{2} v^{\mu}d2vμ. Recall that any given component, for example, v 3 v 3 v^(3)v^{3}v3, is an ordinary scalar function of position (as contrasted to v v v\boldsymbol{v}v or e 3 e 3 e_(3)\boldsymbol{e}_{3}e3 or e 3 v 3 e 3 v 3 e_(3)v^(3)\boldsymbol{e}_{3} v^{3}e3v3 ). Therefore the standard exterior derivative (Chapter 4) as applied to a scalar function is all that d d d\boldsymbol{d}d can mean in d 2 v μ d 2 v μ d^(2)v^(mu)\boldsymbol{d}^{2} v^{\mu}d2vμ. But for the standard exterior derivative applied twice, one has automatically d 2 v μ = 0 d 2 v μ = 0 d^(2)v^(mu)=0\boldsymbol{d}^{2} v^{\mu}=0d2vμ=0 (Box 4.1, B; Box 4.4). This circumstance reduces the expansion for d 2 v d 2 v d^(2)v\boldsymbol{d}^{2} \boldsymbol{v}d2v to the form
(14.17) d 2 v = e μ μ ν v ν , (14.17) d 2 v = e μ μ ν v ν , {:(14.17)d^(2)v=e_(mu)ℜ^(mu)_(nu)v^(nu)",":}\begin{equation*} \boldsymbol{d}^{2} \boldsymbol{v}=\boldsymbol{e}_{\mu} \Re^{\mu}{ }_{\nu} v^{\nu}, \tag{14.17} \end{equation*}(14.17)d2v=eμμνvν,
where the R μ ν R μ ν R^(mu)_(nu)\mathscr{R}^{\mu}{ }_{\nu}Rμν are abbreviations for the curvature 2-forms
(14.18) R ν μ d ω ν μ + ω α μ ω ν α (14.18) R ν μ d ω ν μ + ω α μ ω ν α {:(14.18)R_(nu)^(mu)-=domega_(nu)^(mu)+omega_(alpha)^(mu)^^omega_(nu)^(alpha):}\begin{equation*} \mathscr{R}_{\nu}^{\mu} \equiv \boldsymbol{d} \boldsymbol{\omega}_{\nu}^{\mu}+\boldsymbol{\omega}_{\alpha}^{\mu} \wedge \boldsymbol{\omega}_{\nu}^{\alpha} \tag{14.18} \end{equation*}(14.18)Rνμdωνμ+ωαμωνα
Ordinarily, equation (14.18) surpasses in efficiency every other known method for calculating the curvature 2 -forms.
The remarkable form of equation (14.17) deserves comment. On the left appear two d d d\boldsymbol{d}d 's, reminders that one has twice differentiated the vector field v v v\boldsymbol{v}v. But on the right, as the result of the differentiation, one has only the vector field v v v\boldsymbol{v}v at the point in question, undifferentiated. How v v v\boldsymbol{v}v varies from place to place enters not one whit in the answer. All that matters is how the geometry varies from place to place. Here is curvature coming into evidence. It comes into evidence free of any special features of the vector field v v v\boldsymbol{v}v, because the operation d 2 d 2 d^(2)\boldsymbol{d}^{2}d2 is an antisymmetrized covariant derivative [compare equation (11.8) for this antisymmetrized covariant derivative in the previously developed abstract language, and see Boxes 11.2 and 11.6 for what is going on behind the scene expressed in the form of pictures]. In brief, the result of operating on v v v\boldsymbol{v}v twice with d d d\boldsymbol{d}d is an algebraic linear operation on v v v\boldsymbol{v}v; thus,
(14.19) d 2 v = R v (14.19) d 2 v = R v {:(14.19)d^(2)v=R^(⏜)v:}\begin{equation*} \boldsymbol{d}^{2} \boldsymbol{v}=\overparen{R} \mathbf{v} \tag{14.19} \end{equation*}(14.19)d2v=Rv
Here R R R\mathscr{R}R is an abbreviation for the " ( 1 1 ) ( 1 1 ) ((1)/(1))\binom{1}{1}(11)-tensor valued 2 -form,"
(14.20) R = e μ ω ν R μ ν . (14.20) R = e μ ω ν R μ ν . {:(14.20)R=e_(mu)oxomega^(nu)R^(mu)_(nu).:}\begin{equation*} \mathscr{R}=\boldsymbol{e}_{\mu} \otimes \boldsymbol{\omega}^{\nu} \mathscr{R}^{\mu}{ }_{\nu} . \tag{14.20} \end{equation*}(14.20)R=eμωνRμν.
If d d d\boldsymbol{d}d is a derivative with a "slot in it" in which to insert the vector saying in what direction the differentiation is to proceed, then the d 2 w d 2 w d^(2)w\boldsymbol{d}^{2} \boldsymbol{w}d2w of d 2 w = R w d 2 w = R w d^(2)w=Rw\boldsymbol{d}^{2} \boldsymbol{w}=\mathscr{R} \boldsymbol{w}d2w=Rw has two slots and calls for two vectors, say, u u u\boldsymbol{u}u and v v v\boldsymbol{v}v. These two vectors define the plane in which the antisymmetrized exterior derivative of (14.19) is to be evaluated (change in w w w\boldsymbol{w}w upon going around the elementary route defined by u u u\boldsymbol{u}u and v v v\boldsymbol{v}v and coming back to its starting point; Boxes 11.6 and 11.7). To spell out explicitly this insertion of vectors into slots, return first to a simpler context, and see the exterior derivative of a 1 -form (itself a 2-form) "evaluated" for a bivector u v u v u^^v\boldsymbol{u} \wedge \boldsymbol{v}uv ("count of honeycomblike cells of the 2 -form over the parallelogram-shaped domain defined by the two vectors u u u\boldsymbol{u}u and v " v " v"\boldsymbol{v} "v" ", and see the result of the evaluation (exercise 14.6) expressed as a commutator,
(14.21) d α , u v = u α , v v α , u α , [ u , v ] . (14.21) d α , u v = u α , v v α , u α , [ u , v ] . {:(14.21)(:d alpha","u^^v:)=del_(u)(:alpha","v:)-del_(v)(:alpha","u:)-(:alpha","[u","v]:).:}\begin{equation*} \langle\boldsymbol{d} \boldsymbol{\alpha}, \boldsymbol{u} \wedge \boldsymbol{v}\rangle=\partial_{u}\langle\boldsymbol{\alpha}, \boldsymbol{v}\rangle-\partial_{\boldsymbol{v}}\langle\boldsymbol{\alpha}, \boldsymbol{u}\rangle-\langle\boldsymbol{\alpha},[\boldsymbol{u}, \boldsymbol{v}]\rangle . \tag{14.21} \end{equation*}(14.21)dα,uv=uα,vvα,uα,[u,v].
Tensor-valued curvature 2-form \wp
This result generalizes itself to a tensor-valued 1-form S S S\boldsymbol{S}S of any rank in an obvious way; thus,
(14.22) d S , u v = u S , v v S , u S , [ u , v ] . (14.22) d S , u v = u S , v v S , u S , [ u , v ] . {:(14.22)(:dS","u^^v:)=grad_(u)(:S","v:)-grad_(v)(:S","u:)-(:S","[u","v]:).:}\begin{equation*} \langle\boldsymbol{d} S, \boldsymbol{u} \wedge \boldsymbol{v}\rangle=\boldsymbol{\nabla}_{u}\langle\boldsymbol{S}, \boldsymbol{v}\rangle-\nabla_{v}\langle\boldsymbol{S}, \boldsymbol{u}\rangle-\langle\boldsymbol{S},[\boldsymbol{u}, \boldsymbol{v}]\rangle . \tag{14.22} \end{equation*}(14.22)dS,uv=uS,vvS,uS,[u,v].
Apply this result to the vector-valued 1 -form S = d w S = d w S=dw\boldsymbol{S}=\boldsymbol{d} \boldsymbol{w}S=dw. Recall the expression for a directional derivative, d w , u = u w d w , u = u w (:dw,u:)=grad_(u)w\langle\boldsymbol{d} \boldsymbol{w}, \boldsymbol{u}\rangle=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{w}dw,u=uw. Thus find the result
(14.23) d 2 w , u v = u v w v u w [ u , v ] w = R ( u , v ) w , (14.23) d 2 w , u v = u v w v u w [ u , v ] w = R ( u , v ) w , {:[(14.23)(:d^(2)w,u^^v:)=grad_(u)grad_(v)w-grad_(v)grad_(u)w-grad_([u,v])w],[=R(u","v)w","]:}\begin{align*} \left\langle d^{2} w, u \wedge v\right\rangle & =\nabla_{u} \nabla_{v} w-\nabla_{v} \nabla_{u} w-\nabla_{[u, v]} w \tag{14.23}\\ & =\mathscr{R}(u, v) w, \end{align*}(14.23)d2w,uv=uvwvuw[u,v]w=R(u,v)w,
where R ( u , v ) R ( u , v ) R(u,v)\mathscr{R}(\boldsymbol{u}, \boldsymbol{v})R(u,v) is the curvature operator defined already in Chapter 11 [equation (11.8)]. The conclusion is simple: the ( 1 1 ) ( 1 1 ) ((1)/(1))\binom{1}{1}(11)-tensor-valued 2 -form \mathscr{\Re} of (14.19), evaluated on the bivector ("parallelogram") u v u v u^^v\boldsymbol{u} \wedge \boldsymbol{v}uv, is identical with the curvature operator R ( u , v ) R ( u , v ) R(u,v)\mathscr{R}(\boldsymbol{u}, \boldsymbol{v})R(u,v) introduced previously; thus
(14.24) R , u v = R ( u , v ) (14.24) R , u v = R ( u , v ) {:(14.24)(:R","u^^v:)=R(u","v):}\begin{equation*} \langle\mathscr{R}, \boldsymbol{u} \wedge \boldsymbol{v}\rangle=\mathscr{R}(\boldsymbol{u}, \boldsymbol{v}) \tag{14.24} \end{equation*}(14.24)R,uv=R(u,v)
Now go from the language of abstract operators to a language that begins to make components show up. Substitute on the left the expression (14.20) and on the right the value of the curvature operator from (11.11); and rewrite (14.24) in the form
e μ ω ν R μ p , u v = e μ ω ν R μ ν α β u α v β e μ ω ν R μ p , u v = e μ ω ν R μ ν α β u α v β e_(mu)oxomega^(nu)(:R^(mu)_(p),u^^v:)=e_(mu)oxomega^(nu)R^(mu)_(nu alpha beta)u^(alpha)v^(beta)\boldsymbol{e}_{\mu} \otimes \boldsymbol{\omega}^{\nu}\left\langle\mathscr{R}^{\mu}{ }_{p}, \boldsymbol{u} \wedge \boldsymbol{v}\right\rangle=\boldsymbol{e}_{\mu} \otimes \boldsymbol{\omega}^{\nu} R^{\mu}{ }_{\nu \alpha \beta} u^{\alpha} v^{\beta}eμωνRμp,uv=eμωνRμναβuαvβ
Compare and conclude that the typical individual curvature 2 -form is given by the formula
(14.25) R μ ν = R μ p | α β | ω α ω β (14.25) R μ ν = R μ p | α β | ω α ω β {:(14.25)R^(mu)_(nu)=R^(mu)_(p|alpha beta|)omega^(alpha)^^omega^(beta):}\begin{equation*} \mathscr{R}^{\mu}{ }_{\nu}=R^{\mu}{ }_{p|\alpha \beta|} \boldsymbol{\omega}^{\alpha} \wedge \boldsymbol{\omega}^{\beta} \tag{14.25} \end{equation*}(14.25)Rμν=Rμp|αβ|ωαωβ
(sum over α , β α , β alpha,beta\alpha, \betaα,β, restricted to α < β α < β alpha < beta\alpha<\betaα<β; so each index pair occurs only once).
Equation (14.25) provides the promised packaging of 21 curvature components into six curvature 2 -forms; and equation (14.18) provides the quick means to calculate these curvature 2 -forms. It is not necessary to take the key calculational equations (14.18) on faith, or to master the extended exterior derivative to prove or use them. Not one mention of any d d d\boldsymbol{d}d do they make except the standard exterior d d d\boldsymbol{d}d of Chapter 4. These key equations, moreover, can be verified in detail (exercise 14.8) by working in a coordinate frame. One adopts basis 1 -forms ω α = d x α ω α = d x α omega^(alpha)=dx^(alpha)\boldsymbol{\omega}^{\alpha}=\boldsymbol{d} x^{\alpha}ωα=dxα. One goes on to use ω μ v = Γ μ p λ d x λ ω μ v = Γ μ p λ d x λ omega^(mu)_(v)=Gamma^(mu)_(p lambda)dx^(lambda)\boldsymbol{\omega}^{\mu}{ }_{v}=\Gamma^{\mu}{ }_{p \lambda} \boldsymbol{d} x^{\lambda}ωμv=Γμpλdxλ from equation (14.15). In this way one obtains the "standard formula for the curvature" [equation (11.12) and equation (3) of Box 14.2] by standard methods.
In summary, the calculus of forms and exterior derivatives reduces the
Γ μ α β R μ ν α β Γ μ α β R μ ν α β Gamma^(mu)_(alpha beta)longrightarrowR^(mu)_(nu alpha beta)\Gamma^{\mu}{ }_{\alpha \beta} \longrightarrow R^{\mu}{ }_{\nu \alpha \beta}ΓμαβRμναβ
calculation to the
ω μ p R μ v ω μ p R μ v omega^(mu)_(p)longrightarrowR^(mu)_(v)\boldsymbol{\omega}^{\mu}{ }_{p} \longrightarrow \mathscr{R}^{\mu}{ }_{v}ωμpRμv
computation. Now look at the other link in the chain that leads from metric to curvature. It used to be
g μ ν Γ α β μ g μ ν Γ α β μ g_(mu nu)longrightarrowGamma_(alpha beta)^(mu)g_{\mu \nu} \longrightarrow \Gamma_{\alpha \beta}^{\mu}gμνΓαβμ
It now reduces to the calculation of "connection 1-forms"; thus
g μ ν ω μ ν g μ ν ω μ ν g_(mu nu)longrightarrowomega^(mu)_(nu)g_{\mu \nu} \longrightarrow \boldsymbol{\omega}^{\mu}{ }_{\nu}gμνωμν
Two principles master this first step in the curvature computation: (1) the symmetry of the covariant derivative; and (2) its compatibility with the metric. Condition (1), symmetry, appears in hidden guise in the principle
(14.26) d 2 P = 0 . (14.26) d 2 P = 0 . {:(14.26)d^(2P)=0.:}\begin{equation*} \boldsymbol{d}^{2 \mathscr{P}}=0 . \tag{14.26} \end{equation*}(14.26)d2P=0.
Here the notation " P P P\mathscr{P}P for point" comes straight out of Cartan. He thought of a vector as defined by the movement of one point to another point infinitesimally close to it. To write d P d P dP\boldsymbol{d} \mathscr{P}dP was therefore to take the "derivative of a point" [make a construction with a "point deleted" (tail of vector) and "point reinserted nearby" (tip of vector)]. The direction of the derivative d d d\boldsymbol{d}d in d P d P dP\boldsymbol{d} \mathscr{P}dP is indefinite. In other words, d P d P dP\boldsymbol{d} \mathscr{P}dP contains a "slot." Only when one inserts into this slot a definite vector v v v\boldsymbol{v}v does d P d P dP\boldsymbol{d} \mathscr{P}dP give a definite answer for Cartan's vector. What is that vector that d P d P dP\boldsymbol{d} \mathscr{P}dP then gives? It is v v v\boldsymbol{v}v itself. "The movement that is v v v\boldsymbol{v}v tells the point P P P\mathscr{P}P to reproduce the movement that is v v v\boldsymbol{v}v "; or in concrete notation,
(14.27) d P , v = v (14.27) d P , v = v {:(14.27)(:dP","v:)=v:}\begin{equation*} \langle\boldsymbol{d} \mathscr{\mathscr { P }}, \boldsymbol{v}\rangle=\boldsymbol{v} \tag{14.27} \end{equation*}(14.27)dP,v=v
Put the content of this equation into more formalistic terms. The quantity d P d P dP\boldsymbol{d} \mathscr{P}dP is a ( 1 1)-tensor
(14.28) d P = e μ ω μ (14.28) d P = e μ ω μ {:(14.28)dP=e_(mu)omega^(mu):}\begin{equation*} \boldsymbol{d} \mathscr{P}=\boldsymbol{e}_{\mu} \boldsymbol{\omega}^{\mu} \tag{14.28} \end{equation*}(14.28)dP=eμωμ
It is distinguished from the generic ( 1 1 ) 1 1 ((1)/(1))\left(\frac{1}{1}\right)(11)-tensor
T = e μ T μ ν ω ν T = e μ T μ ν ω ν T=e_(mu)T^(mu)_(nu)omega^(nu)\boldsymbol{T}=\boldsymbol{e}_{\mu} T^{\mu}{ }_{\nu} \boldsymbol{\omega}^{\nu}T=eμTμνων
by the special value of its components
T μ ν = δ μ ν T μ ν = δ μ ν T^(mu)_(nu)=delta^(mu)_(nu)T^{\mu}{ }_{\nu}=\delta^{\mu}{ }_{\nu}Tμν=δμν
In this sense it deserves the name of "unit tensor." Insert this tensor in place of S S S\boldsymbol{S}S into equation (14.22) and obtain the result
(14.29) d 2 P , u v = u v v u [ u , v ] = 0 (14.29) d 2 P , u v = u v v u [ u , v ] = 0 {:(14.29)(:d^(2)P,u^^v:)=grad_(u)v-grad_(v)u-[u","v]=0:}\begin{equation*} \left\langle\boldsymbol{d}^{2} \mathscr{P}, \boldsymbol{u} \wedge \boldsymbol{v}\right\rangle=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}-[\boldsymbol{u}, \boldsymbol{v}]=0 \tag{14.29} \end{equation*}(14.29)d2P,uv=uvvu[u,v]=0
The zero on the right is a restatement of equation (10.2a) or of "the closing of the vector diagram" in the picture called "symmetry of covariant differentiation" in Box 10.2. The vanishing of the righthand side for arbitrary u u u\boldsymbol{u}u and v v v\boldsymbol{v}v demands the vanishing of d 2 P d 2 P d^(2P)\boldsymbol{d}^{2 \mathscr{P}}d2P on the left; and conversely, the vanishing of d 2 P d 2 P d^(2)P\boldsymbol{d}^{2} \mathscr{P}d2P demands the symmetry of the covariant derivative. The other principle basic to the forthcoming computations is "compatibility of covariant derivative with metric," as expressed in the form
(14.30) d ( u v ) = ( d u ) v + u ( d v ) (14.30) d ( u v ) = ( d u ) v + u ( d v ) {:(14.30)d(u*v)=(du)*v+u*(dv):}\begin{equation*} d(u \cdot v)=(d u) \cdot v+u \cdot(d v) \tag{14.30} \end{equation*}(14.30)d(uv)=(du)v+u(dv)
It is essential here to ascribe to the metric (the "dot") a vanishing covariant derivative; thus
d ( ) = 0 . d ( ) = 0 . d(*)=0.\boldsymbol{d}(\cdot)=0 .d()=0.
Symmetry of covariant derivative:
(1) expressed as d 2 P = 0 d 2 P = 0 d^(2P)=0\boldsymbol{d}^{2 \mathscr{P}}=0d2P=0
Capitalize on the symmetry and compatibility of the covariant derivative by using basis vectors (and where appropriate the basis 1 -forms dual to these basis vectors) in equations (14.26) and (14.30). Thus from
d P = e μ ω μ d P = e μ ω μ dP=e_(mu)omega^(mu)\boldsymbol{d} \mathscr{P}=\boldsymbol{e}_{\mu} \boldsymbol{\omega}^{\mu}dP=eμωμ
compute
0 = d 2 P = d e μ ω μ + e μ d ω μ = e μ ( ω μ ν ω ν + d ω μ ) , 0 = d 2 P = d e μ ω μ + e μ d ω μ = e μ ω μ ν ω ν + d ω μ , {:[0=d^(2P)=de_(mu)^^omega^(mu)+e_(mu)domega^(mu)],[=e_(mu)(omega^(mu)_(nu)^^omega^(nu)+domega^(mu))","]:}\begin{aligned} 0 & =\boldsymbol{d}^{2 \mathscr{P}}=\boldsymbol{d} \boldsymbol{e}_{\mu} \wedge \boldsymbol{\omega}^{\mu}+\boldsymbol{e}_{\mu} \boldsymbol{d} \boldsymbol{\omega}^{\mu} \\ & =\boldsymbol{e}_{\mu}\left(\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{\omega}^{\nu}+\boldsymbol{d} \boldsymbol{\omega}^{\mu}\right), \end{aligned}0=d2P=deμωμ+eμdωμ=eμ(ωμνων+dωμ),
and conclude that the coefficient of e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ must vanish; or
(2) expressed as
d ω μ + ω μ ν ω ν = 0 d ω μ + ω μ ν ω ν = 0 domega^(mu)+omega^(mu)_(nu)^^omega^(nu)=0\boldsymbol{d} \boldsymbol{\omega}^{\mu}+\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{\omega}^{\nu}=0dωμ+ωμνων=0
Compatibility of g g g\boldsymbol{g}g and grad\boldsymbol{\nabla} expressed as d g μ ν = ω μ ν + ω ν μ d g μ ν = ω μ ν + ω ν μ dg_(mu nu)=omega_(mu nu)+omega_(nu mu)\boldsymbol{d} g_{\mu \nu}=\boldsymbol{\omega}_{\mu \nu}+\boldsymbol{\omega}_{\nu \mu}dgμν=ωμν+ωνμ
(14.31a) 0 = d ω μ + ω μ v ω ν ("symmetry") (14.31a) 0 = d ω μ + ω μ v ω ν  ("symmetry")  {:(14.31a)0=domega^(mu)+omega^(mu)_(v)^^omega^(nu)quad" ("symmetry") ":}\begin{equation*} 0=\boldsymbol{d} \boldsymbol{\omega}^{\mu}+\boldsymbol{\omega}^{\mu}{ }_{v} \wedge \boldsymbol{\omega}^{\nu} \quad \text { ("symmetry") } \tag{14.31a} \end{equation*}(14.31a)0=dωμ+ωμvων ("symmetry") 
Next, into (14.30) in place of the general u u u\boldsymbol{u}u and v v v\boldsymbol{v}v insert the specific e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ and e ν e ν e_(nu)\boldsymbol{e}_{\nu}eν, respectively, and find
(14.31b) d g μ ν = ω μ ν + ω ν μ ("compatibility"), (14.31b) d g μ ν = ω μ ν + ω ν μ  ("compatibility"),  {:(14.31b)dg_(mu nu)=omega_(mu nu)+omega_(nu mu)quad" ("compatibility"), ":}\begin{equation*} \boldsymbol{d} g_{\mu \nu}=\boldsymbol{\omega}_{\mu \nu}+\boldsymbol{\omega}_{\nu \mu} \quad \text { ("compatibility"), } \tag{14.31b} \end{equation*}(14.31b)dgμν=ωμν+ωνμ ("compatibility"), 
where
(14.31c) ω μ ν g μ α ω α ν = Γ μ ν α ω α . (14.31c) ω μ ν g μ α ω α ν = Γ μ ν α ω α . {:(14.31c)omega_(mu nu)-=g_(mu alpha)omega^(alpha)_(nu)=Gamma_(mu nu alpha)omega^(alpha).:}\begin{equation*} \boldsymbol{\omega}_{\mu \nu} \equiv g_{\mu \alpha} \boldsymbol{\omega}^{\alpha}{ }_{\nu}=\Gamma_{\mu \nu \alpha} \boldsymbol{\omega}^{\alpha} . \tag{14.31c} \end{equation*}(14.31c)ωμνgμαωαν=Γμναωα.
In equations (14.31) one has the connection between metric and connection forms expressed in the most compact way.

§14.6. COMPUTATION OF CURVATURE USING EXTERIOR DIFFERENTIAL FORMS

The use of differential forms for the computation of curvature is illustrated in Box 14.5. This section outlines the method. There are three main steps: compute ω μ v ω μ v omega^(mu)_(v)\boldsymbol{\omega}^{\mu}{ }_{v}ωμv; compute μ v μ v ℜ^(mu)_(v)\Re^{\mu}{ }_{v}μv; and compute G μ v G μ v G^(mu)_(v)G^{\mu}{ }_{v}Gμv. More particularly, first select a metric and a frame. Thereby fix the basis forms ω μ = L α μ d x α ω μ = L α μ d x α omega^(mu)=L_(alpha^('))^(mu)dx^(alpha^('))\boldsymbol{\omega}^{\mu}=L_{\alpha^{\prime}}^{\mu} \boldsymbol{d} x^{\alpha^{\prime}}ωμ=Lαμdxα and the metric components g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν in d s 2 = g μ ν ω μ ω ν d s 2 = g μ ν ω μ ω ν ds^(2)=g_(mu nu)omega^(mu)oxomega^(nu)\boldsymbol{d} \boldsymbol{s}^{2}=g_{\mu \nu} \boldsymbol{\omega}^{\mu} \otimes \boldsymbol{\omega}^{\nu}ds2=gμνωμων. Then determine the connection forms ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν, and determine them uniquely, as solutions of the equations
(14.31a) 0 = d ω μ + ω ν μ ω ν , (14.31b) d g μ ν = ω μ ν + ω ν μ . (14.31a) 0 = d ω μ + ω ν μ ω ν , (14.31b) d g μ ν = ω μ ν + ω ν μ . {:[(14.31a)0=domega^(mu)+omega_(nu)^(mu)^^omega^(nu)","],[(14.31b)dg_(mu nu)=omega_(mu nu)+omega_(nu mu).]:}\begin{align*} 0 & =\boldsymbol{d} \boldsymbol{\omega}^{\mu}+\boldsymbol{\omega}_{\nu}^{\mu} \wedge \boldsymbol{\omega}^{\nu}, \tag{14.31a}\\ \boldsymbol{d} g_{\mu \nu} & =\boldsymbol{\omega}_{\mu \nu}+\boldsymbol{\omega}_{\nu \mu} . \tag{14.31b} \end{align*}(14.31a)0=dωμ+ωνμων,(14.31b)dgμν=ωμν+ωνμ.
The "guess and check" method of finding a solution to these equations (described and illustrated in Box 14.5) is often quick and easy. [Exercise (14.7) shows that a solution always exists by showing that the Christoffel formula (14.36) is the unique solution in coordinate frames.] It is usually most convenient to use an orthonormal frame with g μ ν = η μ ν g μ ν = η μ ν g_(mu nu)=eta_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}gμν=ημν (or some other simple frame where g μ ν = g μ ν = g_(mu nu)=g_{\mu \nu}=gμν= const, e.g., a null frame). Then d g μ ν = 0 d g μ ν = 0 dg_(mu nu)=0\boldsymbol{d} g_{\mu \nu}=0dgμν=0 and equation (14.31b) shows that ω μ ν = ω ν μ ω μ ν = ω ν μ omega_(mu nu)=-omega_(nu mu)\boldsymbol{\omega}_{\mu \nu}=-\boldsymbol{\omega}_{\nu \mu}ωμν=ωνμ. Therefore there are only six ω μ ν ω μ ν omega_(mu nu)\boldsymbol{\omega}_{\mu \nu}ωμν for which to solve in four dimensions.

Box 14.5 CURVATURE COMPUTED USING EXTERIOR DIFFERENTIAL FORMS (METRIC FOR FRIEDMANN COSMOLOGY)

The Friedmann metric
d s 2 = d t 2 + a 2 ( t ) [ d χ 2 + sin 2 χ ( d θ 2 + sin 2 θ d ϕ 2 ) ] d s 2 = d t 2 + a 2 ( t ) d χ 2 + sin 2 χ d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-dt^(2)+a^(2)(t)[dchi^(2)+sin^(2)chi(dtheta^(2)+sin^(2)theta dphi^(2))]d s^{2}=-d t^{2}+a^{2}(t)\left[d \chi^{2}+\sin ^{2} \chi\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right]ds2=dt2+a2(t)[dχ2+sin2χ(dθ2+sin2θdϕ2)]
(Box 27.1) represents a spacetime where each constant- t t ttt hypersurface is a threedimensional hypersphere of proper circumference 2 π a ( t ) 2 π a ( t ) 2pi a(t)2 \pi a(t)2πa(t). An orthonormal basis is easily found in this spacetime; thus,
d s 2 = ( ω t ^ ) 2 + ( ω χ ^ ) 2 + ( ω θ ^ ) 2 + ( ω ϕ ^ ) 2 d s 2 = ω t ^ 2 + ω χ ^ 2 + ω θ ^ 2 + ω ϕ ^ 2 ds^(2)=-(omega^( hat(t)))^(2)+(omega^( hat(chi)))^(2)+(omega^( hat(theta)))^(2)+(omega^( hat(phi)))^(2)\boldsymbol{d} \boldsymbol{s}^{2}=-\left(\boldsymbol{\omega}^{\hat{t}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\chi}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\theta}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\phi}}\right)^{2}ds2=(ωt^)2+(ωχ^)2+(ωθ^)2+(ωϕ^)2
where
ω t ^ = d t , ω χ ^ = a d χ , (1) ω θ ^ = a sin χ d θ , ω ϕ ^ = a sin χ sin θ d ϕ . ω t ^ = d t , ω χ ^ = a d χ , (1) ω θ ^ = a sin χ d θ , ω ϕ ^ = a sin χ sin θ d ϕ . {:[omega^( hat(t))=dt","],[omega^( hat(chi))=ad chi","],[(1)omega^( hat(theta))=a sin chi d theta","],[omega^( hat(phi))=a sin chi sin theta d phi.]:}\begin{align*} \boldsymbol{\omega}^{\hat{t}} & =\boldsymbol{d} t, \\ \boldsymbol{\omega}^{\hat{\chi}} & =a \boldsymbol{d} \chi, \\ \boldsymbol{\omega}^{\hat{\theta}} & =a \sin \chi \boldsymbol{d} \theta, \tag{1}\\ \boldsymbol{\omega}^{\hat{\phi}} & =a \sin \chi \sin \theta \boldsymbol{d} \phi . \end{align*}ωt^=dt,ωχ^=adχ,(1)ωθ^=asinχdθ,ωϕ^=asinχsinθdϕ.

A. Connection Computation

Equation (14.31b) gives, since d g μ ν = d η μ ν = 0 d g μ ν = d η μ ν = 0 dg_(mu nu)=deta_(mu nu)=0\boldsymbol{d} g_{\mu \nu}=\boldsymbol{d} \eta_{\mu \nu}=0dgμν=dημν=0, just
(2) ω μ ν = ω ν μ ; (2) ω μ ν = ω ν μ ; {:(2)omega_(mu nu)=-omega_(nu mu);:}\begin{equation*} \boldsymbol{\omega}_{\mu \nu}=-\boldsymbol{\omega}_{\nu \mu} ; \tag{2} \end{equation*}(2)ωμν=ωνμ;
so there are only six 1-forms ω μ ν ω μ ν omega_(mu nu)\boldsymbol{\omega}_{\mu \nu}ωμν to be found. Turn to the second basic equation (14.31a). The game now is to guess a solution (because this is so often quicker than using systematic methods) to the equations 0 = d ω μ + ω μ ν ω ν 0 = d ω μ + ω μ ν ω ν 0=domega^(mu)+omega^(mu)_(nu)^^omega^(nu)0=\boldsymbol{d} \boldsymbol{\omega}^{\mu}+\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{\omega}^{\nu}0=dωμ+ωμνων in which the ω ν ω ν omega^(nu)\boldsymbol{\omega}^{\nu}ων and thus also d ω μ d ω μ domega^(mu)\boldsymbol{d} \boldsymbol{\omega}^{\mu}dωμ are known, and ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν are unknown. The solution ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν is known to be unique; so guessing (if it leads to any answer) can only give the right answer.
Proceed from the simplest such equation. From ω t ^ = d t ω t ^ = d t omega^( hat(t))=dt\boldsymbol{\omega}^{\hat{t}}=\boldsymbol{d} tωt^=dt, compute
d ω t ^ = 0 d ω t ^ = 0 domega^( hat(t))=0\boldsymbol{d} \boldsymbol{\omega}^{\hat{t}}=0dωt^=0
Compare this with d ω t ^ = ω t ^ μ ω μ d ω t ^ = ω t ^ μ ω μ domega^( hat(t))=-omega^( hat(t))_(mu)^^omega^(mu)\boldsymbol{d} \boldsymbol{\omega}^{\hat{t}}=-\boldsymbol{\omega}^{\hat{t}}{ }_{\mu} \wedge \boldsymbol{\omega}^{\mu}dωt^=ωt^μωμ or (since ω t ^ t ^ = ω i ^ t ^ = 0 ω t ^ t ^ = ω i ^ t ^ = 0 omega_( hat(t))^( hat(t))=-omega_( hat(i) hat(t))=0\boldsymbol{\omega}_{\hat{t}}^{\hat{t}}=-\boldsymbol{\omega}_{\hat{i} \hat{t}}=0ωt^t^=ωi^t^=0, by ω μ ν = ω ν μ ω μ ν = ω ν μ omega_(mu nu)=-omega_(nu mu)\boldsymbol{\omega}_{\mu \nu}=-\boldsymbol{\omega}_{\nu \mu}ωμν=ωνμ )
d ω t ^ = ω k t ^ ω k = 0 d ω t ^ = ω k t ^ ω k = 0 domega^( hat(t))=-omega_(k)^( hat(t))^^omega^(k)=0\boldsymbol{d} \boldsymbol{\omega}^{\hat{t}}=-\boldsymbol{\omega}_{k}^{\hat{t}} \wedge \boldsymbol{\omega}^{k}=0dωt^=ωkt^ωk=0
This equation could be satisfied by having ω t ^ k ω k ω t ^ k ω k omega^( hat(t))_(k)propomega^(k)\boldsymbol{\omega}^{\hat{t}}{ }_{k} \propto \boldsymbol{\omega}^{k}ωt^kωk, or in more complicated ways with cancelations among different terms, or more simply by ω t ^ k = 0 ω t ^ k = 0 omega^( hat(t))_(k)=0\boldsymbol{\omega}^{\hat{t}}{ }_{k}=0ωt^k=0. Proceed, not

Box 14.5 (continued)

looking for trouble, until some non-zero ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν is required. From ω x ^ = a d χ ω x ^ = a d χ omega^( hat(x))=ad chi\boldsymbol{\omega}^{\hat{x}}=a \boldsymbol{d} \chiωx^=adχ, find
d ω x ^ = a ˙ d t d X = ( a ˙ / a ) ω t ^ ω x ^ = ( a ˙ / a ) ω x ^ ω t ^ . d ω x ^ = a ˙ d t d X = ( a ˙ / a ) ω t ^ ω x ^ = ( a ˙ / a ) ω x ^ ω t ^ . {:[domega^( hat(x))=a^(˙)dt^^dX],[=(a^(˙)//a)omega^( hat(t))^^omega^( hat(x))=-(a^(˙)//a)omega^( hat(x))^^omega^( hat(t)).]:}\begin{aligned} \boldsymbol{d} \boldsymbol{\omega}^{\hat{x}} & =\dot{a} \boldsymbol{d} t \wedge \boldsymbol{d} X \\ & =(\dot{a} / a) \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{x}}=-(\dot{a} / a) \boldsymbol{\omega}^{\hat{x}} \wedge \boldsymbol{\omega}^{\hat{t}} . \end{aligned}dωx^=a˙dtdX=(a˙/a)ωt^ωx^=(a˙/a)ωx^ωt^.
Compare this with
d ω x ^ = ω x ^ μ ω μ = ω x ^ t ^ ω t ^ ω x ^ θ ^ ω θ ^ ω x ^ ϕ ^ ω ϕ ^ . d ω x ^ = ω x ^ μ ω μ = ω x ^ t ^ ω t ^ ω x ^ θ ^ ω θ ^ ω x ^ ϕ ^ ω ϕ ^ . {:[domega^( hat(x))=-omega^( hat(x)_(mu))^^omega^(mu)],[=-omega^( hat(x)_( hat(t)))^^omega^( hat(t))-omega^( hat(x)_( hat(theta)))^^omega^( hat(theta))-omega^( hat(x)_( hat(phi)))^^omega^( hat(phi)).]:}\begin{aligned} \boldsymbol{d} \boldsymbol{\omega}^{\hat{x}} & =-\boldsymbol{\omega}^{\hat{x}_{\mu}} \wedge \boldsymbol{\omega}^{\mu} \\ & =-\boldsymbol{\omega}^{\hat{x}_{\hat{t}}} \wedge \boldsymbol{\omega}^{\hat{t}}-\boldsymbol{\omega}^{\hat{x}_{\hat{\theta}}} \wedge \boldsymbol{\omega}^{\hat{\theta}}-\boldsymbol{\omega}^{\hat{x}_{\hat{\phi}}} \wedge \boldsymbol{\omega}^{\hat{\phi}} . \end{aligned}dωx^=ωx^μωμ=ωx^t^ωt^ωx^θ^ωθ^ωx^ϕ^ωϕ^.
Guess that ω χ ^ t ^ = ( a ˙ / a ) ω χ ^ ω χ ^ t ^ = ( a ˙ / a ) ω χ ^ omega^( hat(chi)_( hat(t)))=(a^(˙)//a)omega hat(chi)\boldsymbol{\omega}^{\hat{\chi}_{\hat{t}}}=(\dot{a} / a) \boldsymbol{\omega} \hat{\chi}ωχ^t^=(a˙/a)ωχ^ from the first term; and hope the other terms vanish. (Note that this allows ω t ^ χ ^ ω χ ^ = ω t ^ χ ^ ω χ ^ = ω χ ^ t ^ ω χ ^ = 0 ω t ^ χ ^ ω χ ^ = ω t ^ χ ^ ω χ ^ = ω χ ^ t ^ ω χ ^ = 0 omega^( hat(t)) hat(chi)^^omega^( hat(chi))=-omega_( hat(t) hat(chi))^^omega^( hat(chi))=omega_( hat(chi) hat(t))^^omega^( hat(chi))=0\boldsymbol{\omega}^{\hat{t}} \hat{\chi} \wedge \boldsymbol{\omega}^{\hat{\chi}}=-\boldsymbol{\omega}_{\hat{t} \hat{\chi}} \wedge \boldsymbol{\omega}^{\hat{\chi}}=\boldsymbol{\omega}_{\hat{\chi} \hat{t}} \wedge \boldsymbol{\omega}^{\hat{\chi}}=0ωt^χ^ωχ^=ωt^χ^ωχ^=ωχ^t^ωχ^=0 in the d ω t ^ d ω t ^ domega^( hat(t))\boldsymbol{d} \boldsymbol{\omega}^{\hat{t}}dωt^ equation.) Look at ω θ ^ = a sin χ d θ ω θ ^ = a sin χ d θ omega^( hat(theta))=a sin chi d theta\boldsymbol{\omega}^{\hat{\theta}}=a \sin \chi \boldsymbol{d} \thetaωθ^=asinχdθ, and write
d ω θ ^ ( a ˙ / a ) ω t ^ ω θ ^ + a 1 cot χ ω χ ^ ω θ ^ = ω t ^ θ ^ ω t ^ ω θ ^ χ ^ ω x ^ ω θ ^ ϕ ^ ω ϕ ^ . d ω θ ^ ( a ˙ / a ) ω t ^ ω θ ^ + a 1 cot χ ω χ ^ ω θ ^ = ω t ^ θ ^ ω t ^ ω θ ^ χ ^ ω x ^ ω θ ^ ϕ ^ ω ϕ ^ . {:[domega^( hat(theta))(a^(˙)//a)omega^( hat(t))^^omega^( hat(theta))+a^(-1)cot chiomega^( hat(chi))^^omega^( hat(theta))],[=-omega_( hat(t))^( hat(theta))^^omega^( hat(t))-omega^( hat(theta)) hat(chi)^^omega^( hat(x))-omega^( hat(theta))_( hat(phi))^^omega^( hat(phi)).]:}\begin{aligned} & \boldsymbol{d} \boldsymbol{\omega}^{\hat{\theta}}(\dot{a} / a) \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{\theta}}+a^{-1} \cot \chi \boldsymbol{\omega}^{\hat{\chi}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \\ & =-\boldsymbol{\omega}_{\hat{t}}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{t}}-\boldsymbol{\omega}^{\hat{\theta}} \hat{\chi} \wedge \boldsymbol{\omega}^{\hat{x}}-\boldsymbol{\omega}^{\hat{\theta}}{ }_{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{\phi}} . \end{aligned}dωθ^(a˙/a)ωt^ωθ^+a1cotχωχ^ωθ^=ωt^θ^ωt^ωθ^χ^ωx^ωθ^ϕ^ωϕ^.
Guess, consistent with previously written equations, that
ω θ ^ = ω t ^ θ ^ θ ^ ω θ ^ χ ^ ^ = ω x ^ θ ^ = a 1 cot χ ω θ ^ . ω θ ^ = ω t ^ θ ^ θ ^ ω θ ^ χ ^ ^ = ω x ^ θ ^ = a 1 cot χ ω θ ^ . {:[omega^( hat(theta))=omega^( hat(t)) hat(theta)_( hat(theta))],[omega^( hat(theta)) hat(hat(chi))=-omega^( hat(x)_( hat(theta)))=a^(-1)cot chiomega^( hat(theta)).]:}\begin{aligned} \boldsymbol{\omega}^{\hat{\theta}} & =\boldsymbol{\omega}^{\hat{t}} \hat{\theta}_{\hat{\theta}} \\ \boldsymbol{\omega}^{\hat{\theta}} \hat{\hat{\chi}} & =-\boldsymbol{\omega}^{\hat{x}_{\hat{\theta}}}=a^{-1} \cot \chi \boldsymbol{\omega}^{\hat{\theta}} . \end{aligned}ωθ^=ωt^θ^θ^ωθ^χ^^=ωx^θ^=a1cotχωθ^.
Finally from
d ω ϕ ^ = ( a ˙ / a ) ω t ^ ω ϕ ^ + a 1 cot χ ω χ ^ ω ϕ ^ + ( a sin χ ) 1 cot θ ω θ ^ ω ϕ ^ = ω ϕ ^ t ^ ω t ^ ω χ ^ χ ^ ω χ ^ ω ϕ ^ θ ^ ω θ ^ , d ω ϕ ^ = ( a ˙ / a ) ω t ^ ω ϕ ^ + a 1 cot χ ω χ ^ ω ϕ ^ + ( a sin χ ) 1 cot θ ω θ ^ ω ϕ ^ = ω ϕ ^ t ^ ω t ^ ω χ ^ χ ^ ω χ ^ ω ϕ ^ θ ^ ω θ ^ , {:[domega^( hat(phi))=(a^(˙)//a)omega^( hat(t))^^omega^( hat(phi))+a^(-1)cot chiomega^( hat(chi))^^omega^( hat(phi))],[+(a sin chi)^(-1)cot thetaomega^( hat(theta))^^omega^( hat(phi))],[=-omega^( hat(phi)) hat(t)^^omega^( hat(t))-omega^( hat(chi)) hat(chi)^^omega^( hat(chi))-omega^( hat(phi))_( hat(theta))^^omega^( hat(theta))","]:}\begin{aligned} \boldsymbol{d} \boldsymbol{\omega}^{\hat{\phi}}= & (\dot{a} / a) \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{\phi}}+a^{-1} \cot \chi \boldsymbol{\omega}^{\hat{\chi}} \wedge \boldsymbol{\omega}^{\hat{\phi}} \\ & +(a \sin \chi)^{-1} \cot \theta \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\phi}} \\ = & -\boldsymbol{\omega}^{\hat{\phi}} \hat{t} \wedge \boldsymbol{\omega}^{\hat{t}}-\boldsymbol{\omega}^{\hat{\chi}} \hat{\chi} \wedge \boldsymbol{\omega}^{\hat{\chi}}-\boldsymbol{\omega}^{\hat{\phi}}{ }_{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\theta}}, \end{aligned}dωϕ^=(a˙/a)ωt^ωϕ^+a1cotχωχ^ωϕ^+(asinχ)1cotθωθ^ωϕ^=ωϕ^t^ωt^ωχ^χ^ωχ^ωϕ^θ^ωθ^,
deduce values of ω ϕ ^ t ^ ^ , ω ϕ ^ χ ^ ω ϕ ^ t ^ ^ , ω ϕ ^ χ ^ omega^( hat(phi)) hat(hat(t)),omega^( hat(phi)_( hat(chi)))\boldsymbol{\omega}^{\hat{\phi}} \hat{\hat{t}}, \boldsymbol{\omega}^{\hat{\phi}_{\hat{\chi}}}ωϕ^t^^,ωϕ^χ^, and ω ϕ ^ θ ^ ϕ ^ ω ϕ ^ θ ^ ϕ ^ omega^( hat(phi)_( hat(theta))^( hat(phi)))\boldsymbol{\omega}^{\hat{\phi}_{\hat{\boldsymbol{\theta}}}^{\hat{\phi}}}ωϕ^θ^ϕ^. These are not inconsistent with previous assumptions that terms like ω θ ^ ϕ ^ ω ϕ ^ ω θ ^ ϕ ^ ω ϕ ^ omega^( hat(theta))_( hat(phi))^^omega^( hat(phi))\boldsymbol{\omega}^{\hat{\theta}}{ }_{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{\phi}}ωθ^ϕ^ωϕ^ vanish (in the d ω θ ^ d ω θ ^ domega^( hat(theta))\boldsymbol{d} \boldsymbol{\omega}^{\hat{\theta}}dωθ^ equation); so one has in fact solved d ω μ = ω μ ν ω ν d ω μ = ω μ ν ω ν domega^(mu)=-omega^(mu)_(nu)^^omega^(nu)\boldsymbol{d} \boldsymbol{\omega}^{\mu}=-\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{\omega}^{\nu}dωμ=ωμνων for a set of connection forms ω μ p ω μ p omega^(mu)_(p)\boldsymbol{\omega}^{\mu}{ }_{p}ωμp, as follows:
ω k t ^ = ω t ^ k = ( a ˙ / a ) ω k , ω θ ^ ^ χ ^ = ω χ ^ θ ^ = a 1 cot χ ω θ ^ = cos χ d θ , (3) ω ϕ ^ χ ^ = ω χ ^ ϕ ^ = a 1 cot χ ω ϕ ^ = cos χ sin θ d ϕ , ω ϕ ^ θ ^ = ω θ ^ ϕ ^ ϕ ^ = ( a sin χ ) 1 cot θ ω ϕ ^ = cos θ d ϕ . ω k t ^ = ω t ^ k = ( a ˙ / a ) ω k , ω θ ^ ^ χ ^ = ω χ ^ θ ^ = a 1 cot χ ω θ ^ = cos χ d θ , (3) ω ϕ ^ χ ^ = ω χ ^ ϕ ^ = a 1 cot χ ω ϕ ^ = cos χ sin θ d ϕ , ω ϕ ^ θ ^ = ω θ ^ ϕ ^ ϕ ^ = ( a sin χ ) 1 cot θ ω ϕ ^ = cos θ d ϕ . {:[omega^(k)_( hat(t))=omega^( hat(t))_(k)=(a^(˙)//a)omega^(k)","],[omega^( hat(hat(theta))) hat(chi)=-omega^( hat(chi)_( hat(theta)))=a^(-1)cot chiomega^( hat(theta))],[=cos chi d theta","],[(3)omega^( hat(phi)_( hat(chi)))=-omega hat(chi)_( hat(phi))=a^(-1)cot chiomega^( hat(phi))],[=cos chi sin theta d phi","],[omega^( hat(phi))_( hat(theta))=-omega^( hat(theta)) hat(phi)_( hat(phi))=(a sin chi)^(-1)cot thetaomega^( hat(phi))],[=cos theta d phi.]:}\begin{align*} & \boldsymbol{\omega}^{k}{ }_{\hat{t}}=\boldsymbol{\omega}^{\hat{t}}{ }_{k}=(\dot{a} / a) \boldsymbol{\omega}^{k}, \\ & \boldsymbol{\omega}^{\hat{\hat{\theta}}} \hat{\chi}=-\boldsymbol{\omega}^{\hat{\chi}_{\hat{\theta}}}=a^{-1} \cot \chi \boldsymbol{\omega}^{\hat{\theta}} \\ & =\cos \chi \boldsymbol{d} \theta, \\ & \boldsymbol{\omega}^{\hat{\phi}_{\hat{\chi}}}=-\boldsymbol{\omega} \hat{\chi}_{\hat{\phi}}=a^{-1} \cot \chi \boldsymbol{\omega}^{\hat{\phi}} \tag{3}\\ & =\cos \chi \sin \theta \boldsymbol{d} \phi, \\ & \boldsymbol{\omega}^{\hat{\phi}}{ }_{\hat{\theta}}=-\boldsymbol{\omega}^{\hat{\theta}} \hat{\phi}_{\hat{\phi}}=(a \sin \chi)^{-1} \cot \theta \boldsymbol{\omega}^{\hat{\phi}} \\ & =\cos \theta \boldsymbol{d} \phi . \end{align*}ωkt^=ωt^k=(a˙/a)ωk,ωθ^^χ^=ωχ^θ^=a1cotχωθ^=cosχdθ,(3)ωϕ^χ^=ωχ^ϕ^=a1cotχωϕ^=cosχsinθdϕ,ωϕ^θ^=ωθ^ϕ^ϕ^=(asinχ)1cotθωϕ^=cosθdϕ.
Of course, if these hit-or-miss methods of finding ω μ p ω μ p omega^(mu)_(p)\boldsymbol{\omega}^{\mu}{ }_{p}ωμp do not work easily in some problem, one may simply use equations (14.32) and (14.33).

B. Curvature Computation

The curvature computation is a straightforward substitution of ω μ v ω μ v omega^(mu)_(v)\boldsymbol{\omega}^{\mu}{ }_{v}ωμv from equations (3) above into equation (14.34), which is
R μ ν = d ω μ ν + ω α μ ω α ν . R μ ν = d ω μ ν + ω α μ ω α ν . R^(mu)_(nu)=domega^(mu)_(nu)+omega_(alpha)^(mu)^^omega^(alpha)_(nu).\mathscr{R}^{\mu}{ }_{\nu}=\boldsymbol{d} \boldsymbol{\omega}^{\mu}{ }_{\nu}+\boldsymbol{\omega}_{\alpha}^{\mu} \wedge \boldsymbol{\omega}^{\alpha}{ }_{\nu} .Rμν=dωμν+ωαμωαν.
This equation is short enough that one can write out the sum
in contrast to the ten terms in the corresponding R = Γ + Γ 2 R = Γ + Γ 2 R=del Gamma+Gamma^(2)R=\partial \Gamma+\Gamma^{2}R=Γ+Γ2 equation [equation (3) of Box 14.2]. Warning!: From ω χ ^ χ ^ = ( a ˙ / a ) ω χ ^ x ^ ω χ ^ χ ^ = ( a ˙ / a ) ω χ ^ x ^ omega_( hat(chi))^( hat(chi))=(a^(˙)//a)omega hat(chi)^( hat(x))\boldsymbol{\omega}_{\hat{\chi}}^{\hat{\chi}}=(\dot{a} / a) \boldsymbol{\omega} \hat{\chi}^{\hat{x}}ωχ^χ^=(a˙/a)ωχ^x^, do not compute d ω x ^ ^ t ^ ^ = d ω x ^ ^ t ^ ^ = domega_( hat(hat(x)))^( hat(hat(t)))=\boldsymbol{d} \boldsymbol{\omega}_{\hat{\hat{x}}}^{\hat{\hat{t}}}=dωx^^t^^= ( a ˙ / a ) ω t ^ ω x ^ ( a ˙ / a ) ω t ^ ω x ^ (a^(˙)//a)^(*)omega^( hat(t))^^omega^( hat(x))(\dot{a} / a)^{\cdot} \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{x}}(a˙/a)ωt^ωx^. Missing is the term ( a ˙ / a ) d ω χ ^ ( a ˙ / a ) d ω χ ^ (a^(˙)//a)domega^( hat(chi))(\dot{a} / a) \boldsymbol{d} \boldsymbol{\omega}^{\hat{\chi}}(a˙/a)dωχ^. Instead write ω t ^ χ ^ ^ = ( a ˙ / a ) ω χ ^ = ω t ^ χ ^ ^ = ( a ˙ / a ) ω χ ^ = omega^( hat(t)) hat(hat(chi))=(a^(˙)//a)omega^( hat(chi))=\boldsymbol{\omega}^{\hat{t}} \hat{\hat{\chi}}=(\dot{a} / a) \boldsymbol{\omega}^{\hat{\chi}}=ωt^χ^^=(a˙/a)ωχ^= a ˙ d χ a ˙ d χ a^(˙)d chi\dot{a} \boldsymbol{d} \chia˙dχ, and then find d ω χ ^ t ^ = a ¨ d t d χ = ( a ¨ / a ) ω t ^ ω χ ^ d ω χ ^ t ^ = a ¨ d t d χ = ( a ¨ / a ) ω t ^ ω χ ^ domega_( hat(chi))^( hat(t))=a^(¨)dt^^d_(chi)=(a^(¨)//a)omega^( hat(t))^^omega^( hat(chi))\boldsymbol{d} \boldsymbol{\omega}_{\hat{\chi}}^{\hat{t}}=\ddot{a} \boldsymbol{d} t \wedge \boldsymbol{d}_{\chi}=(\ddot{a} / a) \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{\chi}}dωχ^t^=a¨dtdχ=(a¨/a)ωt^ωχ^. With elementary care, then, in correctly substituting from (3) for the ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν in the formula for R μ p R μ p R^(mu)_(p)\mathscr{R}^{\mu}{ }_{p}Rμp, one finds
and
R χ ^ χ ^ = ( a ¨ / a ) ω t ^ ω x ^ R χ ^ χ ^ = ( a ¨ / a ) ω t ^ ω x ^ R_( hat(chi))^( hat(chi))=(a^(¨)//a)omega^( hat(t))^^omega^( hat(x))\mathscr{R}_{\hat{\chi}}^{\hat{\chi}}=(\ddot{a} / a) \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{x}}Rχ^χ^=(a¨/a)ωt^ωx^
R x ^ θ ^ = ( 1 + a ˙ 2 ) a 2 ω α ^ ω θ ^ . R x ^ θ ^ = 1 + a ˙ 2 a 2 ω α ^ ω θ ^ . R hat(x)_( hat(theta))=(1+a^(˙)^(2))a^(-2)omega^( hat(alpha))^^omega^( hat(theta)).\mathscr{R} \hat{x}_{\hat{\theta}}=\left(1+\dot{a}^{2}\right) a^{-2} \boldsymbol{\omega}^{\hat{\alpha}} \wedge \boldsymbol{\omega}^{\hat{\theta}} .Rx^θ^=(1+a˙2)a2ωα^ωθ^.
This completes the computation of the R μ v α β R μ v α β R^(mu)_(v alpha beta)R^{\mu}{ }_{v \alpha \beta}Rμvαβ, since in this isotopic model universe, all space directions in the orthonormal frame ω μ ω μ omega^(mu)\boldsymbol{\omega}^{\mu}ωμ are algebraically equivalent. One can therefore write
(4) R k ^ = ( a ~ / a ) ω t ^ ω k , R ι k = a 2 ( 1 + a ˙ 2 ) ω k ω , (4) R k ^ = ( a ~ / a ) ω t ^ ω k , R ι k = a 2 1 + a ˙ 2 ω k ω , {:[(4)R_(k)^( hat())=( tilde(a)//a)omega^( hat(t))^^omega^(k)","],[R_(iota)^(k)=a^(-2)(1+a^(˙)^(2))omega^(k)^^omega^(ℓ)","]:}\begin{align*} \mathscr{R}_{k}^{\hat{}} & =(\tilde{a} / a) \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{k}, \tag{4}\\ \mathscr{R}_{\iota}^{k} & =a^{-2}\left(1+\dot{a}^{2}\right) \boldsymbol{\omega}^{k} \wedge \boldsymbol{\omega}^{\ell}, \end{align*}(4)Rk^=(a~/a)ωt^ωk,Rιk=a2(1+a˙2)ωkω,
for the complete list of R μ ν R μ ν R^(mu)_(nu)\mathscr{R}^{\mu}{ }_{\nu}Rμν. Specific components, such as
R t χ ^ x ^ x ^ = a ¨ / a , R t ^ χ ^ t ^ θ ^ = 0 , etc. , R t χ ^ x ^ x ^ = a ¨ / a , R t ^ χ ^ t ^ θ ^ = 0 ,  etc.  , R^(t_( hat(chi) hat(x) hat(x))=a^(¨)//a,quadR^( hat(t)_( hat(chi) hat(t) hat(theta)))=0," etc. ",)R^{t_{\hat{\chi} \hat{x} \hat{x}}=\ddot{a} / a, \quad R^{\hat{t}_{\hat{\chi} \hat{t} \hat{\theta}}}=0, \text { etc. }, ~}Rtχ^x^x^=a¨/a,Rt^χ^t^θ^=0, etc. , 
or
R t ^ θ ^ χ ^ ϕ ^ = 0 , R ϕ ^ ϕ ^ θ ^ ϕ ^ = a 2 ( 1 + a ˙ 2 ) R t ^ θ ^ χ ^ ϕ ^ = 0 , R ϕ ^ ϕ ^ θ ^ ϕ ^ = a 2 1 + a ˙ 2 R^( hat(t) hat(theta) hat(chi) hat(phi))=0,quadR_( hat(phi) hat(phi)_( hat(theta) hat(phi)))=a^(-2)(1+a^(˙)^(2))R^{\hat{t} \hat{\theta} \hat{\chi} \hat{\phi}}=0, \quad R_{\hat{\phi} \hat{\phi}_{\hat{\theta} \hat{\phi}}}=a^{-2}\left(1+\dot{a}^{2}\right)Rt^θ^χ^ϕ^=0,Rϕ^ϕ^θ^ϕ^=a2(1+a˙2)
are easily read out of this display of R μ ν R μ ν R^(mu)_(nu)\mathscr{R}{ }^{\mu}{ }_{\nu}Rμν.

C. Contraction

From equations (14.7), find
(5a) G f t ^ = + 3 a 2 ( 1 + a ˙ 2 ) , (5b) G x ^ ^ = G θ ^ ^ θ ^ = G ϕ ^ ^ p ^ = 0 = G x ^ θ ^ = G θ ^ ϕ ^ = G ϕ ^ x ^ , (5c) G x ^ x ^ = G θ ^ θ ^ = G ϕ ^ ϕ ^ = [ 2 a 1 a ¨ + a 2 ( 1 + a ˙ 2 ) ] , (5a) G f t ^ = + 3 a 2 1 + a ˙ 2 , (5b) G x ^ ^ = G θ ^ ^ θ ^ = G ϕ ^ ^ p ^ = 0 = G x ^ θ ^ = G θ ^ ϕ ^ = G ϕ ^ x ^ , (5c) G x ^ x ^ = G θ ^ θ ^ = G ϕ ^ ϕ ^ = 2 a 1 a ¨ + a 2 1 + a ˙ 2 , {:[(5a)G^(f hat(t))=+3a^(-2)(1+a^(˙)^(2))","],[(5b)G^( hat(hat(x)))=G^( hat(hat(theta)) hat(theta))=G^( hat(hat(phi)) hat(p))=0=G^( hat(x) hat(theta))=G^( hat(theta) hat(phi))=G^( hat(phi) hat(x))","],[(5c)G^( hat(x) hat(x))=G^( hat(theta) hat(theta))=G^( hat(phi) hat(phi))=-[2a^(-1)(a^(¨))+a^(-2)(1+a^(˙)^(2))]","]:}\begin{align*} & G^{f \hat{t}}=+3 a^{-2}\left(1+\dot{a}^{2}\right), \tag{5a}\\ & G^{\hat{\hat{x}}}=G^{\hat{\hat{\theta}} \hat{\theta}}=G^{\hat{\hat{\phi}} \hat{p}}=0=G^{\hat{x} \hat{\theta}}=G^{\hat{\theta} \hat{\phi}}=G^{\hat{\phi} \hat{x}}, \tag{5b}\\ & G^{\hat{x} \hat{x}}=G^{\hat{\theta} \hat{\theta}}=G^{\hat{\phi} \hat{\phi}}=-\left[2 a^{-1} \ddot{a}+a^{-2}\left(1+\dot{a}^{2}\right)\right], \tag{5c} \end{align*}(5a)Gft^=+3a2(1+a˙2),(5b)Gx^^=Gθ^^θ^=Gϕ^^p^=0=Gx^θ^=Gθ^ϕ^=Gϕ^x^,(5c)Gx^x^=Gθ^θ^=Gϕ^ϕ^=[2a1a¨+a2(1+a˙2)],
and
(6) R = G μ μ = 6 [ a 1 a ¨ + a 2 ( 1 + a ˙ 2 ) ] (6) R = G μ μ = 6 a 1 a ¨ + a 2 1 + a ˙ 2 {:(6)R=-G^(mu)_(mu)=6[a^(-1)(a^(¨))+a^(-2)(1+a^(˙)^(2))]:}\begin{equation*} R=-G^{\mu}{ }_{\mu}=6\left[a^{-1} \ddot{a}+a^{-2}\left(1+\dot{a}^{2}\right)\right] \tag{6} \end{equation*}(6)R=Gμμ=6[a1a¨+a2(1+a˙2)]
If guessing is not easy, there is a systematic way to solve equations (14.31) in an orthonormal frame or in any other frame in which d g μ ν = 0 d g μ ν = 0 dg_(mu nu)=0\boldsymbol{d} g_{\mu \nu}=0dgμν=0. Compute the d ω μ d ω μ domega^(mu)\boldsymbol{d} \boldsymbol{\omega}^{\mu}dωμ and arrange them in the format
(14.32) d ω α = c | μ ν | α ω μ ω ν . (14.32) d ω α = c | μ ν | α ω μ ω ν . {:(14.32)domega^(alpha)=-c_(|mu nu|)^(alpha)omega^(mu)^^omega^(nu).:}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\alpha}=-c_{|\mu \nu|}{ }^{\alpha} \boldsymbol{\omega}^{\mu} \wedge \boldsymbol{\omega}^{\nu} . \tag{14.32} \end{equation*}(14.32)dωα=c|μν|αωμων.
In this way display the 24 "commutation coefficients" c μ ν α c μ ν α c_(mu nu)^(alpha)c_{\mu \nu}{ }^{\alpha}cμνα. These quantities enter into the formula
(14.33) ω μ ν = 1 2 ( c μ ν α + c μ α ν c ν α μ ) ω α (14.33) ω μ ν = 1 2 c μ ν α + c μ α ν c ν α μ ω α {:(14.33)omega_(mu nu)=(1)/(2)(c_(mu nu alpha)+c_(mu alpha nu)-c_(nu alpha mu))omega^(alpha):}\begin{equation*} \boldsymbol{\omega}_{\mu \nu}=\frac{1}{2}\left(c_{\mu \nu \alpha}+c_{\mu \alpha \nu}-c_{\nu \alpha \mu}\right) \boldsymbol{\omega}^{\alpha} \tag{14.33} \end{equation*}(14.33)ωμν=12(cμνα+cμανcναμ)ωα
to provide the six ω μ p ω μ p omega^(mu)_(p)\boldsymbol{\omega}^{\mu}{ }_{p}ωμp (exercise 14.12).
(3) calculate curvature 2-forms R μ p R μ p R^(mu)_(p)\mathscr{R}^{\mu}{ }_{p}Rμp
Once the ω μ ν ω μ ν omega_(mu nu)\boldsymbol{\omega}_{\mu \nu}ωμν are known, one computes the curvature forms R μ ν R μ ν R^(mu)_(nu)\mathscr{R}^{\mu}{ }_{\nu}Rμν (again only six in four dimensions, since R μ ν = ν μ R μ ν = ν μ R^(mu nu)=-ℜ^(nu mu)\mathscr{R}^{\mu \nu}=-\Re^{\nu \mu}Rμν=νμ ) by use of the formula
(14.34) R μ ν = d ω μ ν + ω α μ ω α ν . (14.34) R μ ν = d ω μ ν + ω α μ ω α ν . {:(14.34)R^(mu)_(nu)=domega^(mu)_(nu)+omega_(alpha)^(mu)^^omega^(alpha)_(nu).:}\begin{equation*} \mathscr{R}^{\mu}{ }_{\nu}=\boldsymbol{d} \boldsymbol{\omega}^{\mu}{ }_{\nu}+\boldsymbol{\omega}_{\alpha}^{\mu} \wedge \boldsymbol{\omega}^{\alpha}{ }_{\nu} . \tag{14.34} \end{equation*}(14.34)Rμν=dωμν+ωαμωαν.
(4) calculate components of curvature tensors
Out of this tabulation, one reads the individual components of the curvature tensor by using the identification scheme
(14.35) R μ ν = R μ ν | α β | ω α ω β . (14.35) R μ ν = R μ ν | α β | ω α ω β . {:(14.35)R^(mu nu)=R^(mu nu)_(|alpha beta|)omega^(alpha)^^omega^(beta).:}\begin{equation*} \mathscr{R}^{\mu \nu}=R^{\mu \nu}{ }_{|\alpha \beta|} \boldsymbol{\omega}^{\alpha} \wedge \boldsymbol{\omega}^{\beta} . \tag{14.35} \end{equation*}(14.35)Rμν=Rμν|αβ|ωαωβ.
The Einstein tensor G μ v G μ v G^(mu)_(v)G^{\mu}{ }_{v}Gμv is computed by scanning the R μ ν R μ ν R^(mu nu)\mathscr{R}^{\mu \nu}Rμν display to find the appropriate R μ ν α β R μ ν α β R^(mu nu)_(alpha beta)R^{\mu \nu}{ }_{\alpha \beta}Rμναβ components for use in formulas (14.7).

EXERCISES

Exercise 14.5. EXTERIOR DERIVATIVE OF A PRODUCT OF FORMS

Establish equation (14.13a) by working up recursively from forms of lower order to forms of higher order. [Hints: Recall from equation (4.27) that for a p p ppp-form
α = α μ 1 μ l d μ 1 d x μ p α = α μ 1 μ l d μ 1 d x μ p alpha=alpha_(∣mu_(1)dotsmu_(l))d^(mu_(1))^^dots^^dx^(mu_(p))\boldsymbol{\alpha}=\alpha_{\mid \mu_{1} \ldots \mu_{l}} \boldsymbol{d}^{\mu_{1}} \wedge \ldots \wedge \boldsymbol{d} x^{\mu_{p}}α=αμ1μldμ1dxμp
the exterior derivative is defined by
d α = α | μ 1 , μ 0 | x μ 0 d x μ 0 d x μ 1 d x μ p . d α = α μ 1 , μ 0 x μ 0 d x μ 0 d x μ 1 d x μ p . d alpha=(delalpha_(|mu_(1),mu_(0)|))/(delx^(mu_(0)))dx^(mu_(0))^^dx^(mu_(1))^^dots^^dx^(mu_(p)).\boldsymbol{d} \alpha=\frac{\partial \boldsymbol{\alpha}_{\left|\mu_{1}, \mu_{0}\right|}}{\partial x^{\mu_{0}}} \boldsymbol{d} x^{\mu_{0}} \wedge \boldsymbol{d} x^{\mu_{1}} \wedge \ldots \wedge \boldsymbol{d} x^{\mu_{p}} .dα=α|μ1,μ0|xμ0dxμ0dxμ1dxμp.
Applied to the product α β α β alpha^^beta\boldsymbol{\alpha} \wedge \boldsymbol{\beta}αβ of two 1 -forms, this formula gives
d ( α β ) = d [ ( α λ d x λ ) ( β μ d x μ ) ] = d [ ( α λ β μ ) ( d x λ d x μ ) ] = ( α λ β μ ) x κ d x κ d x λ d x μ = ( α λ x κ d x κ d x λ ) β μ d x μ ( a λ d x λ ) ( β μ x κ d x κ d x μ ) = ( d ) β α d β . d ( α β ) = d α λ d x λ β μ d x μ = d α λ β μ d x λ d x μ = α λ β μ x κ d x κ d x λ d x μ = α λ x κ d x κ d x λ β μ d x μ a λ d x λ β μ x κ d x κ d x μ = ( d ) β α d β . {:[d(alpha^^beta)=d[(alpha_(lambda)dx^(lambda))^^(beta_(mu)dx^(mu))]],[=d[(alpha_(lambda)beta_(mu))(dx^(lambda)^^dx^(mu))]],[=(del(alpha_(lambda)beta_(mu)))/(delx^(kappa))dx^(kappa)^^dx^(lambda)^^dx^(mu)],[=((delalpha_(lambda))/(delx^(kappa))dx^(kappa)^^dx^(lambda))^^beta_(mu)dx^(mu)-(a_(lambda)dx^(lambda))^^((delbeta_(mu))/(delx^(kappa))dx^(kappa)^^dx^(mu))],[=(d)^^beta-alpha^^d beta.]:}\begin{aligned} \boldsymbol{d}(\boldsymbol{\alpha} \wedge \boldsymbol{\beta}) & =\boldsymbol{d}\left[\left(\alpha_{\lambda} \boldsymbol{d} x^{\lambda}\right) \wedge\left(\beta_{\mu} \boldsymbol{d} x^{\mu}\right)\right] \\ & =\boldsymbol{d}\left[\left(\alpha_{\lambda} \beta_{\mu}\right)\left(\boldsymbol{d} x^{\lambda} \wedge \boldsymbol{d} x^{\mu}\right)\right] \\ & =\frac{\partial\left(\alpha_{\lambda} \beta_{\mu}\right)}{\partial x^{\kappa}} \boldsymbol{d} x^{\kappa} \wedge \boldsymbol{d} x^{\lambda} \wedge \boldsymbol{d} x^{\mu} \\ & =\left(\frac{\partial \alpha_{\lambda}}{\partial x^{\kappa}} \boldsymbol{d} x^{\kappa} \wedge \boldsymbol{d} x^{\lambda}\right) \wedge \beta_{\mu} \boldsymbol{d} x^{\mu}-\left(a_{\lambda} \boldsymbol{d} x^{\lambda}\right) \wedge\left(\frac{\partial \beta_{\mu}}{\partial x^{\kappa}} \boldsymbol{d} x^{\kappa} \wedge \boldsymbol{d} x^{\mu}\right) \\ & =(\boldsymbol{d}) \wedge \boldsymbol{\beta}-\boldsymbol{\alpha} \wedge \boldsymbol{d} \boldsymbol{\beta} . \end{aligned}d(αβ)=d[(αλdxλ)(βμdxμ)]=d[(αλβμ)(dxλdxμ)]=(αλβμ)xκdxκdxλdxμ=(αλxκdxκdxλ)βμdxμ(aλdxλ)(βμxκdxκdxμ)=(d)βαdβ.
Extend the reasoning to forms of higher order.]

Exercise 14.6. RELATIONSHIP BETWEEN EXTERIOR DERIVATIVE AND COMMUTATOR

Establish formula (14.21) by showing (a) that the righthand side is an algebraic linear function of u u u\boldsymbol{u}u and an algebraic linear function of v v v\boldsymbol{v}v, and (b) that the equation holds when u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are coordinate basis vectors u = / x k , v = / x u = / x k , v = / x u=del//delx^(k),v=del//delx^(ℓ)\boldsymbol{u}=\partial / \partial x^{k}, \boldsymbol{v}=\partial / \partial x^{\ell}u=/xk,v=/x.
Exercise 14.7. CHRISTOFFEL FORMULA DERIVED FROM CONNECTION FORMS
In a coordinate frame ω μ = d x μ ω μ = d x μ omega^(mu)=dx^(mu)\boldsymbol{\omega}^{\mu}=\boldsymbol{d} x^{\mu}ωμ=dxμ, show that equation (14.31a) requires Γ μ α β = Γ μ β α Γ μ α β = Γ μ β α Gamma^(mu)_(alpha beta)=Gamma^(mu)_(beta alpha)\Gamma^{\mu}{ }_{\alpha \beta}=\Gamma^{\mu}{ }_{\beta \alpha}Γμαβ=Γμβα, and that, with this symmetry established, equation (14.31b) gives an expression for g μ ν / x α g μ ν / x α delg_(mu nu)//delx^(alpha)\partial g_{\mu \nu} / \partial x^{\alpha}gμν/xα which can be solved to give the Christoffel formula
(14.36) Γ α β μ = 1 2 g μ v ( g v α x β + g ν β x α g α β x v ) (14.36) Γ α β μ = 1 2 g μ v g v α x β + g ν β x α g α β x v {:(14.36)Gamma_(alpha beta)^(mu)=(1)/(2)g^(mu v)((delg_(v alpha))/(delx^(beta))+(delg_(nu beta))/(delx^(alpha))-(delg_(alpha beta))/(delx^(v))):}\begin{equation*} \Gamma_{\alpha \beta}^{\mu}=\frac{1}{2} g^{\mu v}\left(\frac{\partial g_{v \alpha}}{\partial x^{\beta}}+\frac{\partial g_{\nu \beta}}{\partial x^{\alpha}}-\frac{\partial g_{\alpha \beta}}{\partial x^{v}}\right) \tag{14.36} \end{equation*}(14.36)Γαβμ=12gμv(gvαxβ+gνβxαgαβxv)
Substitute ω μ ν = Γ μ ν λ d x λ ω μ ν = Γ μ ν λ d x λ omega^(mu)_(nu)=Gamma^(mu)_(nu lambda)dx^(lambda)\boldsymbol{\omega}^{\mu}{ }_{\nu}=\Gamma^{\mu}{ }_{\nu \lambda} \boldsymbol{d} x^{\lambda}ωμν=Γμνλdxλ into equation (14.18), and from the result read out, according to equation (14.25), the classical formula (3) of Box 14.2 for the components R μ ν α β R μ ν α β R^(mu)_(nu alpha beta)R^{\mu}{ }_{\nu \alpha \beta}Rμναβ.

Exercise 14.9. MATRIX NOTATION FOR REVIEW OF CARTAN STRUCTURE EQUATIONS

Let e ( e 1 , , e n ) e e 1 , , e n e-=(e_(1),dots,e_(n))e \equiv\left(\boldsymbol{e}_{1}, \ldots, \boldsymbol{e}_{n}\right)e(e1,,en) be a row matrix whose entries are the basis vectors, and let ω ω omega\omegaω be a column of basis 1 -forms ω μ ω μ omega^(mu)\boldsymbol{\omega}^{\mu}ωμ. Similarly let Ω = ω μ ν Ω = ω μ ν Omega=||omega^(mu)_(nu)||\Omega=\left\|\boldsymbol{\omega}^{\mu}{ }_{\nu}\right\|Ω=ωμν and R = R μ ν R = R μ ν R=||R^(mu)_(nu)||\mathscr{R}=\left\|\mathscr{R}^{\mu}{ }_{\nu}\right\|R=Rμν be square matrices with 1 -form and 2-form entries. This gives a compact notation in which d e μ = e ν ω ν μ d e μ = e ν ω ν μ de_(mu)=e_(nu)omega^(nu)_(mu)\boldsymbol{d} \boldsymbol{e}_{\mu}=\boldsymbol{e}_{\nu} \boldsymbol{\omega}^{\nu}{ }_{\mu}deμ=eνωνμ and d P = e μ ω μ d P = e μ ω μ dP=e_(mu)omega^(mu)\boldsymbol{d} \mathscr{P}=\boldsymbol{e}_{\mu} \boldsymbol{\omega}^{\mu}dP=eμωμ read
(14.37) d e = e Ω and d P = e ω (14.37) d e = e Ω  and  d P = e ω {:(14.37)de=e Omega" and "dP=e omega:}\begin{equation*} \boldsymbol{d} e=e \Omega \text { and } \boldsymbol{d} \mathscr{P}=e \omega \tag{14.37} \end{equation*}(14.37)de=eΩ and dP=eω
respectively.
(a) From equations (14.37) and d 2 P = 0 d 2 P = 0 d^(2P)=0\boldsymbol{d}^{2 \mathscr{P}}=0d2P=0, derive equation (14.31a) in the form
(14.38) 0 = d ω + Ω ω (14.38) 0 = d ω + Ω ω {:(14.38)0=d omega+Omega^^omega:}\begin{equation*} 0=\boldsymbol{d} \omega+\Omega \wedge \omega \tag{14.38} \end{equation*}(14.38)0=dω+Ωω
[Solution: d 2 P = d e ω + e d ω = e ( Ω ω + d ω ) d 2 P = d e ω + e d ω = e ( Ω ω + d ω ) d^(2)P=de^^omega+ed omega=e(Omega^^omega+d omega)\boldsymbol{d}^{2} \mathscr{P}=\boldsymbol{d} e \wedge \omega+e \boldsymbol{d} \omega=e(\Omega \wedge \omega+\boldsymbol{d} \omega)d2P=deω+edω=e(Ωω+dω).]
(b) Compute d 2 e d 2 e d^(2)e\boldsymbol{d}^{2} ed2e as motivation for definition (14.18), which reads
(14.39) = d Ω + Ω Ω (14.39) = d Ω + Ω Ω {:(14.39)ℜ=d Omega+Omega^^Omega:}\begin{equation*} \Re=\boldsymbol{d} \Omega+\Omega \wedge \Omega \tag{14.39} \end{equation*}(14.39)=dΩ+ΩΩ
(c) From d 2 ω = 0 d 2 ω = 0 d^(2)omega=0\boldsymbol{d}^{2} \omega=0d2ω=0, deduce R ω = 0 R ω = 0 R^^omega=0\mathscr{R} \wedge \omega=0Rω=0 and then decompress the notation to get the antisymmetry relation R [ α β γ ] μ = 0 R [ α β γ ] μ = 0 R_([alpha beta gamma])^(mu)=0R_{[\alpha \beta \gamma]}^{\mu}=0R[αβγ]μ=0.
(d) Compute d R d R dR\boldsymbol{d} \mathscr{R}dR from equation (14.39), and relate it to the Bianchi identity R μ ν [ α β ; γ ] = 0 R μ ν [ α β ; γ ] = 0 R^(mu)_(nu[alpha beta;gamma])=0R^{\mu}{ }_{\nu[\alpha \beta ; \gamma]}=0Rμν[αβ;γ]=0.
(e) Let v = { v μ } v = v μ v={v^(mu)}v=\left\{v^{\mu}\right\}v={vμ} be a column of functions; so v = e v = e μ v μ v = e v = e μ v μ v=ev=e_(mu)v^(mu)\boldsymbol{v}=e v=\boldsymbol{e}_{\mu} v^{\mu}v=ev=eμvμ is a vector field. Compute, in compact notation, d v d v dv\boldsymbol{d} \boldsymbol{v}dv and d 2 v d 2 v d^(2)v\boldsymbol{d}^{2} \boldsymbol{v}d2v to show d 2 v = e R v d 2 v = e R v d^(2)v=eRv\boldsymbol{d}^{2} \boldsymbol{v}=e \mathscr{R} vd2v=eRv (which is equation 14.17).

Exercise 14.10. TRANSFORMATION RULES FOR CONNECTION FORMS IN COMPACT NOTATION

Using the notation of the previous exercise, write e = e A e = e A e^(')=eAe^{\prime}=e Ae=eA in place of e μ = e ν A ν μ e μ = e ν A ν μ e_(mu^('))=e_(nu)A^(nu)_(mu^('))\boldsymbol{e}_{\mu^{\prime}}=\boldsymbol{e}_{\nu} A^{\nu}{ }_{\mu^{\prime}}eμ=eνAνμ, and similarly ω = A 1 ω ω = A 1 ω omega^(')=A^(-1)omega\omega^{\prime}=A^{-1} \omegaω=A1ω, to represent a change of frame. Show that d P e ω = e ω d P e ω = e ω dP-=e omega=e^(')omega^(')\boldsymbol{d} \mathscr{P} \equiv e \omega=e^{\prime} \omega^{\prime}dPeω=eω. Substitute e = e A e = e A e^(')=eAe^{\prime}=e Ae=eA in d e = e Ω d e = e Ω de^(')=e^(')Omega^(')\boldsymbol{d} e^{\prime}=e^{\prime} \Omega^{\prime}de=eΩ to deduce the transformation law
(14.40) Ω = A 1 Ω A + A 1 d A (14.40) Ω = A 1 Ω A + A 1 d A {:(14.40)Omega^(')=A^(-1)Omega A+A^(-1)dA:}\begin{equation*} \Omega^{\prime}=A^{-1} \Omega A+A^{-1} \boldsymbol{d} A \tag{14.40} \end{equation*}(14.40)Ω=A1ΩA+A1dA
Rewrite this in decompressed notation for coordinate frames with A ν μ = x ν / x μ A ν μ = x ν / x μ A^(nu)_(mu^('))=delx^(nu)//delx^(mu^('))A^{\nu}{ }_{\mu^{\prime}}=\partial x^{\nu} / \partial x^{\mu^{\prime}}Aνμ=xν/xμ as a formula of the form Γ μ α β = ( ? ) Γ μ α β = ( ? ) Gamma^(mu^('))_(alpha^(')beta^('))=(?)\Gamma^{\mu^{\prime}}{ }_{\alpha^{\prime} \beta^{\prime}}=(?)Γμαβ=(?).
Exercise 14.11. SPACE IS FLAT IF THE CURVATURE VANISHES (see §11.5)
If coordinates exist in which all straight lines ( d 2 x μ / d λ 2 = 0 ) d 2 x μ / d λ 2 = 0 (d^(2)x^(mu)//dlambda^(2)=0)\left(d^{2} x^{\mu} / d \lambda^{2}=0\right)(d2xμ/dλ2=0) are geodesics, then one says the space is flat. Evidently all Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ and R μ ν α β R μ ν α β R^(mu)_(nu alpha beta)R^{\mu}{ }_{\nu \alpha \beta}Rμναβ vanish in this case, by equation (14.8) and equation (3) in Box 14.2. Show conversely that, if R = 0 R = 0 R=0\mathscr{R}=0R=0, then such coordinates exist. Use the results of the previous problem to find differential equations for a transformation A A AAA to a basis e e e^(')e^{\prime}e where Ω = 0 Ω = 0 Omega^(')=0\Omega^{\prime}=0Ω=0. What are the conditions for complete integrability of these equations? [Note that d f K = F K ( x , f ) d f K = F K ( x , f ) df_(K)=F_(K)(x,f)\boldsymbol{d} f_{K}=F_{K}(x, f)dfK=FK(x,f) is completely integrable if d 2 f K = 0 d 2 f K = 0 d^(2)f_(K)=0\boldsymbol{d}^{2} f_{K}=0d2fK=0 modulo the original equations.] Why will the basis forms ω μ ω μ omega^(mu^('))\boldsymbol{\omega}^{\mu^{\prime}}ωμ in this new frame be coordinate differentials ω μ = d x μ ω μ = d x μ omega^(mu^('))=dx^(mu^('))\boldsymbol{\omega}^{\mu^{\prime}}=\boldsymbol{d} x^{\mu^{\prime}}ωμ=dxμ ?

Exercise 14.12. SYSTEMATIC COMPUTATION OF CONNECTION FORMS IN ORTHONORMAL FRAMES

Deduce equation (14.32) by applying equation (14.21) to basis vectors, using equations (8.14) to define c μ ν α c μ ν α c_(mu nu)^(alpha)c_{\mu \nu}{ }^{\alpha}cμνα. Then show that, in an orthonormal frame (or any frames with g μ ν = g μ ν = g_(mu nu)=g_{\mu \nu}=gμν= const), equation (14.33) provides a solution of equations (14.31), which define ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν. [Compare also equation (8.24b).]

Exercise 14.13. SCHWARZSCHILD CURVATURE FORMS

Use the obvious orthonormal frame ω t ^ = e ϕ d t , ω ı ^ = e Λ d r , ω θ ^ = r d θ , ω ϕ ^ = r sin θ d ϕ ω t ^ = e ϕ d t , ω ı ^ = e Λ d r , ω θ ^ = r d θ , ω ϕ ^ = r sin θ d ϕ omega^( hat(t))=e^(phi)dt,omega^( hat(ı))=e^(Lambda)dr,omega^( hat(theta))=rd theta,omega^( hat(phi))=r sin theta d phi\boldsymbol{\omega}^{\hat{t}}=e^{\boldsymbol{\phi}} \boldsymbol{d} t, \boldsymbol{\omega}^{\hat{\imath}}=e^{\Lambda} \boldsymbol{d} r, \boldsymbol{\omega}^{\hat{\theta}}=r \boldsymbol{d} \theta, \boldsymbol{\omega}^{\hat{\phi}}=r \sin \theta \boldsymbol{d} \phiωt^=eϕdt,ωı^=eΛdr,ωθ^=rdθ,ωϕ^=rsinθdϕ for the Schwarzschild metric
(14.41) d s 2 = e 2 Φ d t 2 + e 2 Λ d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (14.41) d s 2 = e 2 Φ d t 2 + e 2 Λ d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(14.41)ds^(2)=-e^(2Phi)dt^(2)+e^(2Lambda)dr^(2)+r^(2)(dtheta^(2)+sin^(2)theta dphi^(2)):}\begin{equation*} d s^{2}=-e^{2 \Phi} d t^{2}+e^{2 \Lambda} d r^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) \tag{14.41} \end{equation*}(14.41)ds2=e2Φdt2+e2Λdr2+r2(dθ2+sin2θdϕ2)
in which Φ Φ Phi\PhiΦ and Λ Λ Lambda\LambdaΛ are functions of r r rrr only; and compute the curvature forms R μ ^ ν ^ R μ ^ ν ^ R_( hat(mu))^( hat(nu))\mathscr{R}_{\hat{\mu}}^{\hat{\nu}}Rμ^ν^ and the Einstein tensor G μ ^ v ^ G μ ^ v ^ G^( hat(mu))_( hat(v))G^{\hat{\mu}}{ }_{\hat{\boldsymbol{v}}}Gμ^v^ by the methods of Box 14.5. [Answer: i r ^ = E ω t ^ ω r ^ , i ^ θ ^ = E ¯ ω t ^ i r ^ = E ω t ^ ω r ^ , i ^ θ ^ = E ¯ ω t ^ ℜ^(i hat(r))=Eomega^( hat(t))^^omega^( hat(r)),ℜ^( hat(i) hat(theta))= bar(E)omega^( hat(t))^^\Re^{i \hat{r}}=E \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{r}}, \Re^{\hat{i} \hat{\theta}}=\bar{E} \boldsymbol{\omega}^{\hat{t}} \wedgeir^=Eωt^ωr^,i^θ^=E¯ωt^ ω θ ^ , R t ^ ϕ ^ = E ¯ ω t ^ ω ϕ ^ , θ ^ ϕ ^ = F ω θ ^ ω ϕ ^ , ϕ ^ r ^ = F ¯ ω ϕ ^ ω r ^ , r ^ θ ^ = F ¯ ω r ^ ω θ ^ ω θ ^ , R t ^ ϕ ^ = E ¯ ω t ^ ω ϕ ^ , θ ^ ϕ ^ = F ω θ ^ ω ϕ ^ , ϕ ^ r ^ = F ¯ ω ϕ ^ ω r ^ , r ^ θ ^ = F ¯ ω r ^ ω θ ^ omega^( hat(theta)),R^( hat(t) hat(phi))= bar(E)omega^( hat(t))^^omega^( hat(phi)),ℜ^( hat(theta) hat(phi))=Fomega^( hat(theta))^^omega^( hat(phi)),ℜ^( hat(phi) hat(r))= bar(F)omega^( hat(phi))^^omega^( hat(r)),ℜ^( hat(r) hat(theta))= bar(F)omega^( hat(r))^^omega^( hat(theta))\boldsymbol{\omega}^{\hat{\theta}}, \mathscr{R}{ }^{\hat{t} \hat{\phi}}=\bar{E} \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{\phi}}, \Re^{\hat{\theta} \hat{\phi}}=F \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\phi}}, \Re^{\hat{\phi} \hat{r}}=\bar{F} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{r}}, \Re^{\hat{r} \hat{\theta}}=\bar{F} \boldsymbol{\omega}^{\hat{r}} \wedge \boldsymbol{\omega}^{\hat{\theta}}ωθ^,Rt^ϕ^=E¯ωt^ωϕ^,θ^ϕ^=Fωθ^ωϕ^,ϕ^r^=F¯ωϕ^ωr^,r^θ^=F¯ωr^ωθ^, with
E = e 2 Λ ( Φ + Φ 2 Φ Λ ) (14.42) E ¯ = 1 r e 2 Λ Φ F = 1 r 2 ( 1 e 2 Λ ) F ¯ = 1 r e 2 Λ Λ E = e 2 Λ Φ + Φ 2 Φ Λ (14.42) E ¯ = 1 r e 2 Λ Φ F = 1 r 2 1 e 2 Λ F ¯ = 1 r e 2 Λ Λ {:[E=-e^(-2Lambda)(Phi^('')+Phi^('2)-Phi^(')Lambda^('))],[(14.42) bar(E)=-(1)/(r)e^(-2Lambda)Phi^(')],[F=(1)/(r^(2))(1-e^(-2Lambda))],[ bar(F)=(1)/(r)e^(-2Lambda)Lambda^(')]:}\begin{align*} E & =-e^{-2 \Lambda}\left(\Phi^{\prime \prime}+\Phi^{\prime 2}-\Phi^{\prime} \Lambda^{\prime}\right) \\ \bar{E} & =-\frac{1}{r} e^{-2 \Lambda} \Phi^{\prime} \tag{14.42}\\ F & =\frac{1}{r^{2}}\left(1-e^{-2 \Lambda}\right) \\ \bar{F} & =\frac{1}{r} e^{-2 \Lambda} \Lambda^{\prime} \end{align*}E=e2Λ(Φ+Φ2ΦΛ)(14.42)E¯=1re2ΛΦF=1r2(1e2Λ)F¯=1re2ΛΛ
and then
G i ^ t = ( F + 2 F ¯ ) , G γ ^ r ^ = ( F + 2 E ¯ ) , (14.43) G θ ^ θ ^ = G ϕ ^ ϕ ^ = ( E + E ¯ + F ¯ ) , G r ^ t ^ = G θ ^ t = G ϕ ^ i = 0 = G θ ^ r ˙ = G ϕ ^ r ^ = G ϕ ^ θ ^ G i ^ t = ( F + 2 F ¯ ) , G γ ^ r ^ = ( F + 2 E ¯ ) , (14.43) G θ ^ θ ^ = G ϕ ^ ϕ ^ = ( E + E ¯ + F ¯ ) , G r ^ t ^ = G θ ^ t = G ϕ ^ i = 0 = G θ ^ r ˙ = G ϕ ^ r ^ = G ϕ ^ θ ^ {:[G_( hat(i))^(t)=-(F+2 bar(F))","],[G_( hat(gamma))^( hat(r))=-(F+2 bar(E))","],[(14.43)G_( hat(theta))^( hat(theta))=G_( hat(phi))^( hat(phi))=-(E+ bar(E)+ bar(F))","],[G_( hat(r))^( hat(t))=G_( hat(theta))^(t)=G_( hat(phi))^(i)=0=G_( hat(theta))^(r^(˙))=G_( hat(phi))^( hat(r))=G_( hat(phi))^( hat(theta))*]:}\begin{align*} G_{\hat{i}}^{t} & =-(F+2 \bar{F}), \\ G_{\hat{\gamma}}^{\hat{r}} & =-(F+2 \bar{E}), \\ G_{\hat{\theta}}^{\hat{\theta}} & =G_{\hat{\phi}}^{\hat{\phi}}=-(E+\bar{E}+\bar{F}), \tag{14.43}\\ G_{\hat{r}}^{\hat{t}} & =G_{\hat{\theta}}^{t}=G_{\hat{\phi}}^{i}=0=G_{\hat{\theta}}^{\dot{r}}=G_{\hat{\phi}}^{\hat{r}}=G_{\hat{\phi}}^{\hat{\theta}} \cdot \end{align*}Gi^t=(F+2F¯),Gγ^r^=(F+2E¯),(14.43)Gθ^θ^=Gϕ^ϕ^=(E+E¯+F¯),Gr^t^=Gθ^t=Gϕ^i=0=Gθ^r˙=Gϕ^r^=Gϕ^θ^

Exercise 14.14. MATRIX DISPLAY OF THE RIEMANN-TENSOR COMPONENTS

Use the symmetries of the Riemann tensor to justify displaying its components in an orthonormal frame in the form
where the rows are labeled by index pairs μ ^ ν ^ = 01 , 02 μ ^ ν ^ = 01 , 02 hat(mu) hat(nu)=01,02\hat{\mu} \hat{\nu}=01,02μ^ν^=01,02, etc., as shown; and the columns α ^ β ^ α ^ β ^ hat(alpha) hat(beta)\hat{\alpha} \hat{\beta}α^β^, similarly. Here E , F E , F E,FE, FE,F, and H H HHH are each 3 × 3 3 × 3 3xx33 \times 33×3 matrices with (why?)
(14.45) E = E T , F = F T , trace H = 0 (14.45) E = E T , F = F T ,  trace  H = 0 {:(14.45)E=E^(T)","quad F=F^(T)","quad" trace "H=0:}\begin{equation*} E=E^{T}, \quad F=F^{T}, \quad \text { trace } H=0 \tag{14.45} \end{equation*}(14.45)E=ET,F=FT, trace H=0
where E T E T E^(T)E^{T}ET means the transpose of E E EEE.

Exercise 14.15. RIEMANN MATRIX WITH VANISHING EINSTEIN TENSOR

Show that the empty-space Einstein equations G μ ^ ν ^ = 0 G μ ^ ν ^ = 0 G^( hat(mu)_( hat(nu)))=0G^{\hat{\mu}_{\hat{\nu}}}=0Gμ^ν^=0 allow the matrix in equation (14.44) to be simplified to the form
(14.46) R α ^ β μ ^ v ^ β ^ = ( E H H E ) , (14.46) R α ^ β μ ^ v ^ β ^ = E H H E , {:(14.46)R_( hat(alpha)beta)^( hat(mu) hat(v)_( hat(beta)))=([E,H],[-H,E])",":}R_{\hat{\alpha} \beta}^{\hat{\mu} \hat{v}_{\hat{\beta}}}=\left(\begin{array}{c:c} E & H \tag{14.46}\\ \hdashline-H & E \end{array}\right),(14.46)Rα^βμ^v^β^=(EHHE),
where now, in addition to the equality E = F E = F E=FE=FE=F that this form implies, the further conditions
(14.47) trace E = 0 , H = H T (14.47)  trace  E = 0 , H = H T {:(14.47)" trace "E=0","H=H^(T):}\begin{equation*} \text { trace } E=0, H=H^{T} \tag{14.47} \end{equation*}(14.47) trace E=0,H=HT
hold.

Exercise 14.16. COMPUTATION OF CURVATURE FOR A PULSATING OR COLLAPSING STAR

Spherically symmetric motions of self-gravitating bodies are discussed in Chapters 26 and 32. A metric form often adopted in this situation is
(14.48) d s 2 = e 2 Φ d T 2 + e 2 Λ d R 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (14.48) d s 2 = e 2 Φ d T 2 + e 2 Λ d R 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(14.48)ds^(2)=-e^(2Phi)dT^(2)+e^(2Lambda)dR^(2)+r^(2)(dtheta^(2)+sin^(2)theta dphi^(2)):}\begin{equation*} d s^{2}=-e^{2 \Phi} d T^{2}+e^{2 \Lambda} d R^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) \tag{14.48} \end{equation*}(14.48)ds2=e2ΦdT2+e2ΛdR2+r2(dθ2+sin2θdϕ2)
where now Φ , Λ Φ , Λ Phi,Lambda\Phi, \LambdaΦ,Λ, and r r rrr are each functions of the two coordinates R R RRR and T T TTT. Compute the curvature 2 -forms and the Einstein tensor for this metric, using the methods of Box 14.5 . In the guessing of the ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν, most of the terms will already be evident from the corresponding calculation in exercise 14.13. [Answer, in the obvious orthonormal frame ω T ^ = e ϕ d T , ω k ^ = ω T ^ = e ϕ d T , ω k ^ = omega^( hat(T))=e^(phi)dT,omega^( hat(k))=\boldsymbol{\omega}^{\hat{T}}=e^{\phi} \boldsymbol{d} T, \boldsymbol{\omega}^{\hat{k}}=ωT^=eϕdT,ωk^= e Λ d R , ω θ ^ = r d θ , ω ϕ ^ = r sin θ d ϕ : e Λ d R , ω θ ^ = r d θ , ω ϕ ^ = r sin θ d ϕ : e^(Lambda)dR,omega^( hat(theta))=rd theta,omega^( hat(phi))=r sin theta d phi:e^{\Lambda} \boldsymbol{d} R, \boldsymbol{\omega}^{\hat{\theta}}=r \boldsymbol{d} \theta, \boldsymbol{\omega}^{\hat{\phi}}=r \sin \theta \boldsymbol{d} \phi:eΛdR,ωθ^=rdθ,ωϕ^=rsinθdϕ:
R R ^ = E ω T ^ ω R ^ R θ ^ ^ = E ¯ ω T ^ ω θ ^ + H ω R ^ ω θ ^ (14.49) r ^ ϕ ^ = E ¯ ω T ^ ω ϕ ^ + H ω R ^ ω α ^ R ϕ ^ = F ω θ ^ ω ϕ ^ R R ^ θ ^ = F ¯ ω R ^ ω θ ^ H ω T ^ ω θ ^ R ϕ ^ = F ¯ ω R ^ ω ϕ ^ H ω T ^ ω ϕ ^ R R ^ = E ω T ^ ω R ^ R θ ^ ^ = E ¯ ω T ^ ω θ ^ + H ω R ^ ω θ ^ (14.49) r ^ ϕ ^ = E ¯ ω T ^ ω ϕ ^ + H ω R ^ ω α ^ R ϕ ^ = F ω θ ^ ω ϕ ^ R R ^ θ ^ = F ¯ ω R ^ ω θ ^ H ω T ^ ω θ ^ R ϕ ^ = F ¯ ω R ^ ω ϕ ^ H ω T ^ ω ϕ ^ {:[R_( hat(R))=Eomega^( hat(T))^^omega^( hat(R))],[R_( hat(hat(theta)))= bar(E)omega^( hat(T))^^omega^( hat(theta))+Homega^( hat(R))^^omega^( hat(theta))],[(14.49)ℜ^( hat(r))_( hat(phi))= bar(E)omega^( hat(T))^^omega^( hat(phi))+Homega^( hat(R))^^omega^( hat(alpha))],[R_( hat(phi))=Fomega^( hat(theta))^^omega^( hat(phi))],[R_( hat(R))^( hat(theta))= bar(F)omega^( hat(R))^^omega^( hat(theta))-Homega^( hat(T))^^omega^( hat(theta))],[R_( hat(phi))= bar(F)omega^( hat(R))^^omega^( hat(phi))-Homega^( hat(T))^^omega^( hat(phi))]:}\begin{align*} & \mathscr{R}_{\hat{R}}=E \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{R}} \\ & \mathscr{R}_{\hat{\hat{\theta}}}=\bar{E} \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{\theta}}+H \boldsymbol{\omega}^{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \\ & \Re^{\hat{r}}{ }_{\hat{\phi}}=\bar{E} \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{\phi}}+H \boldsymbol{\omega}^{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{\alpha}} \tag{14.49}\\ & \mathscr{R}_{\hat{\phi}}=F \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\phi}} \\ & \mathscr{R}_{\hat{R}}^{\hat{\theta}}=\bar{F} \boldsymbol{\omega}^{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{\theta}}-H \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \\ & \mathscr{R}_{\hat{\phi}}=\bar{F} \boldsymbol{\omega}^{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{\phi}}-H \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{\phi}} \end{align*}RR^=EωT^ωR^Rθ^^=E¯ωT^ωθ^+HωR^ωθ^(14.49)r^ϕ^=E¯ωT^ωϕ^+HωR^ωα^Rϕ^=Fωθ^ωϕ^RR^θ^=F¯ωR^ωθ^HωT^ωθ^Rϕ^=F¯ωR^ωϕ^HωT^ωϕ^
which, in the matrix display of exercise 14.14 , gives
Here
E = e 2 Φ ( Λ ¨ + Λ ˙ 2 Λ ˙ Φ ˙ ) e 2 Λ ( Φ + Φ 2 Φ Λ ) , E ¯ = 1 r e 2 Φ ( r ¨ r ˙ Φ ˙ ) 1 r e 2 Λ r Φ , (14.51) H = 1 r e Φ Λ ( r ˙ r ˙ Φ r Λ ˙ ) , F = 1 r 2 ( 1 r 2 e 2 Λ + r ˙ 2 e 2 Φ ) , F ¯ = 1 r e 2 ϕ r ˙ Λ ˙ + 1 r e 2 Λ ( r Λ r ) . E = e 2 Φ Λ ¨ + Λ ˙ 2 Λ ˙ Φ ˙ e 2 Λ Φ + Φ 2 Φ Λ , E ¯ = 1 r e 2 Φ ( r ¨ r ˙ Φ ˙ ) 1 r e 2 Λ r Φ , (14.51) H = 1 r e Φ Λ r ˙ r ˙ Φ r Λ ˙ , F = 1 r 2 1 r 2 e 2 Λ + r ˙ 2 e 2 Φ , F ¯ = 1 r e 2 ϕ r ˙ Λ ˙ + 1 r e 2 Λ r Λ r . {:[E=e^(-2Phi)((Lambda^(¨))+Lambda^(˙)^(2)-(Lambda^(˙))(Phi^(˙)))-e^(-2Lambda)(Phi^('')+Phi^('2)-Phi^(')Lambda^('))","],[ bar(E)=(1)/(r)e^(-2Phi)(r^(¨)-r^(˙)Phi^(˙))-(1)/(r)e^(-2Lambdar^(')Phi^('))","],[(14.51)H=(1)/(r)e^(-Phi-Lambda)(r^(˙)^(')-(r^(˙))Phi^(')-r^(')(Lambda^(˙)))","],[F=(1)/(r^(2))(1-r^('2)e^(-2Lambda)+r^(˙)^(2)e^(-2Phi))","],[ bar(F)=(1)/(r)e^(-2phi)r^(˙)Lambda^(˙)+(1)/(r)e^(-2Lambda(r^(')Lambda^(')-r^('')).)]:}\begin{align*} & E=e^{-2 \Phi}\left(\ddot{\Lambda}+\dot{\Lambda}^{2}-\dot{\Lambda} \dot{\Phi}\right)-e^{-2 \Lambda}\left(\Phi^{\prime \prime}+\Phi^{\prime 2}-\Phi^{\prime} \Lambda^{\prime}\right), \\ & \bar{E}=\frac{1}{r} e^{-2 \Phi}(\ddot{r}-\dot{r} \dot{\Phi})-\frac{1}{r} e^{-2 \Lambda r^{\prime} \Phi^{\prime}}, \\ & H=\frac{1}{r} e^{-\Phi-\Lambda}\left(\dot{r}^{\prime}-\dot{r} \Phi^{\prime}-r^{\prime} \dot{\Lambda}\right), \tag{14.51}\\ & F=\frac{1}{r^{2}}\left(1-r^{\prime 2} e^{-2 \Lambda}+\dot{r}^{2} e^{-2 \Phi}\right), \\ & \bar{F}=\frac{1}{r} e^{-2 \phi} \dot{r} \dot{\Lambda}+\frac{1}{r} e^{-2 \Lambda\left(r^{\prime} \Lambda^{\prime}-r^{\prime \prime}\right) .} \end{align*}E=e2Φ(Λ¨+Λ˙2Λ˙Φ˙)e2Λ(Φ+Φ2ΦΛ),E¯=1re2Φ(r¨r˙Φ˙)1re2ΛrΦ,(14.51)H=1reΦΛ(r˙r˙ΦrΛ˙),F=1r2(1r2e2Λ+r˙2e2Φ),F¯=1re2ϕr˙Λ˙+1re2Λ(rΛr).
The Einstein tensor is
G T ^ = G r ^ Y ^ = F + 2 F ¯ , G T ^ = G r ^ R ^ = 2 H , (14.52) G T ^ θ ^ = G T ^ ϕ ^ = 0 , G R ^ R ^ = ( 2 E ¯ + F ) , G θ ^ ^ θ ^ = G ϕ ^ ϕ ^ = ( E + E ¯ + F ¯ ) , G R ^ θ ^ = G R ^ ϕ ^ = G ϕ ^ θ ^ ϕ ^ = 0 . ] G T ^ = G r ^ Y ^ = F + 2 F ¯ , G T ^ = G r ^ R ^ = 2 H , (14.52) G T ^ θ ^ = G T ^ ϕ ^ = 0 , G R ^ R ^ = ( 2 E ¯ + F ) , G θ ^ ^ θ ^ = G ϕ ^ ϕ ^ = ( E + E ¯ + F ¯ ) , G R ^ θ ^ = G R ^ ϕ ^ = G ϕ ^ θ ^ ϕ ^ = 0 . {:[G^( hat(T))=-G^( hat(r)_( hat(Y)))=F+2 bar(F)","],[G^( hat(T))=G^( hat(r)_( hat(R)))=2H","],[(14.52)G^( hat(T)_( hat(theta)))=G^( hat(T)_( hat(phi)))=0","],[G_( hat(R))^( hat(R))=-(2 bar(E)+F)","],[G_( hat(hat(theta)))^( hat(theta))=G^( hat(phi)_( hat(phi)))=-(E+ bar(E)+ bar(F))","],[{:G^( hat(R)_( hat(theta)))=G^( hat(R)_( hat(phi)))=G_( hat(phi))^( hat(theta)_( hat(phi)))=0.]]:}\begin{align*} & G^{\hat{T}}=-G^{\hat{r}_{\hat{Y}}}=F+2 \bar{F}, \\ & G^{\hat{T}}=G^{\hat{r}_{\hat{R}}}=2 H, \\ & G^{\hat{T}_{\hat{\theta}}}=G^{\hat{T}_{\hat{\phi}}}=0, \tag{14.52}\\ & G_{\hat{R}}^{\hat{R}}=-(2 \bar{E}+F), \\ & G_{\hat{\hat{\theta}}}^{\hat{\theta}}=G^{\hat{\phi}_{\hat{\phi}}}=-(E+\bar{E}+\bar{F}), \\ &\left.G^{\hat{R}_{\hat{\theta}}}=G^{\hat{R}_{\hat{\phi}}}=G_{\hat{\phi}}^{\hat{\theta}_{\hat{\phi}}}=0 .\right] \end{align*}GT^=Gr^Y^=F+2F¯,GT^=Gr^R^=2H,(14.52)GT^θ^=GT^ϕ^=0,GR^R^=(2E¯+F),Gθ^^θ^=Gϕ^ϕ^=(E+E¯+F¯),GR^θ^=GR^ϕ^=Gϕ^θ^ϕ^=0.]

Exercise 14.17. BIANCHI IDENTITY IN d R = 0 d R = 0 dR=0d \mathscr{R}=0dR=0 FORM

Define the Riemann tensor as a bivector-valued 2-form,
(14.53) R = 1 2 e μ e ν R μ ν (14.53) R = 1 2 e μ e ν R μ ν {:(14.53)R=(1)/(2)e_(mu)^^e_(nu)R^(mu nu):}\begin{equation*} \mathscr{R}=\frac{1}{2} \boldsymbol{e}_{\mu} \wedge \boldsymbol{e}_{\nu} \mathscr{R}^{\mu \nu} \tag{14.53} \end{equation*}(14.53)R=12eμeνRμν
and evaluate d R d R dR\boldsymbol{d} \mathscr{R}dR to make it manifest that d R = 0 d R = 0 dR=0\boldsymbol{d} \mathscr{R}=0dR=0. Use
(14.54) R μ ν = d ω μ ν ω α μ ω ν α , (14.54) R μ ν = d ω μ ν ω α μ ω ν α , {:(14.54)R^(mu nu)=domega^(mu nu)-omega_(alpha)^(mu)^^omega^(nu alpha)",":}\begin{equation*} \mathscr{R}^{\mu \nu}=\boldsymbol{d} \boldsymbol{\omega}^{\mu \nu}-\boldsymbol{\omega}_{\alpha}^{\mu} \wedge \boldsymbol{\omega}^{\nu \alpha}, \tag{14.54} \end{equation*}(14.54)Rμν=dωμνωαμωνα,
which is derived easily in an orthonormal frame (adequate for proving d R = 0 d R = 0 dR=0\boldsymbol{d} \mathscr{R}=0dR=0 ), or (as a test of skill) in a general frame where R μ ν = R μ α g α ν R μ ν = R μ α g α ν R^(mu nu)=R^(mu)_(alpha)g^(alpha nu)\mathscr{R}^{\mu \nu}=\mathscr{R}^{\mu}{ }_{\alpha} g^{\alpha \nu}Rμν=Rμαgαν and (why?) d g μ ν = g μ α ( d g α β ) g β ν d g μ ν = g μ α d g α β g β ν dg^(mu nu)=-g^(mu alpha)(dg_(alpha beta))g^(beta nu)\boldsymbol{d} g^{\mu \nu}=-g^{\mu \alpha}\left(\boldsymbol{d} g_{\alpha \beta}\right) g^{\beta \nu}dgμν=gμα(dgαβ)gβν. [Note: only wedge products between forms (not those between vectors) count in fixing signs in the product rule (14.13) for d d d\boldsymbol{d}d.]

Exercise 14.18. LOCAL CONSERVATION OF ENERGY AND MOMENTUM: d T = 0 d T = 0 d^(**)T=0\boldsymbol{d}^{*} \boldsymbol{T}=0dT=0 MEANS T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0

Let the duality operator * , as defined for exterior differential forms in Box 4.1, act on the forms, but not on the contravariant vectors, which appear when the stress-energy tensor T T T\boldsymbol{T}T or the Einstein tensor G G G\boldsymbol{G}G is written as a mixed ( 1 1 ) ( 1 1 ) ((1)/(1))\binom{1}{1}(11) tensor:
or
T = e μ T μ ν ω ν T = e μ T μ ν ω ν T=e_(mu)T^(mu)_(nu)omega^(nu)\boldsymbol{T}=\boldsymbol{e}_{\mu} T^{\mu}{ }_{\nu} \boldsymbol{\omega}^{\nu}T=eμTμνων
G = e μ G μ ν ω ν G = e μ G μ ν ω ν G=e_(mu)G^(mu)_(nu)omega^(nu)\boldsymbol{G}=\boldsymbol{e}_{\mu} G^{\mu}{ }_{\nu} \boldsymbol{\omega}^{\nu}G=eμGμνων
(a) Give an expression for T T ^(**)T{ }^{*} \boldsymbol{T}T (or G G ^(**)G{ }^{*} \boldsymbol{G}G ) expanded in terms of basis vectors and forms.
(b) Show that
T = e μ T μ ν d 3 Σ p T = e μ T μ ν d 3 Σ p ^(**)T=e_(mu)T^(mu nu)d^(3)Sigma_(p){ }^{*} \boldsymbol{T}=\boldsymbol{e}_{\mu} T^{\mu \nu} d^{3} \Sigma_{p}T=eμTμνd3Σp
where d 3 Σ p = ϵ p | α β γ | ω α ω β ω γ d 3 Σ p = ϵ p | α β γ | ω α ω β ω γ d^(3)Sigma_(p)=epsilon_(p|alpha beta gamma|)omega^(alpha)^^omega^(beta)^^omega^(gamma)d^{3} \Sigma_{p}=\boldsymbol{\epsilon}_{p|\alpha \beta \gamma|} \boldsymbol{\omega}^{\alpha} \wedge \boldsymbol{\omega}^{\beta} \wedge \boldsymbol{\omega}^{\gamma}d3Σp=ϵp|αβγ|ωαωβωγ [see Box 5.4 and equations (8.10)].
(c) Compute d T d T d^(**)T\boldsymbol{d}^{*} \boldsymbol{T}dT using the generalized exterior derivative d d d\boldsymbol{d}d; find that
d T = e μ T μ ν ; ν | g | ω 0 ω 1 ω 2 ω 3 d T = e μ T μ ν ; ν | g | ω 0 ω 1 ω 2 ω 3 d^(**)T=e_(mu)T^(mu nu)_(;nu)sqrt(|g|)omega^(0)^^omega^(1)^^omega^(2)^^omega^(3)\boldsymbol{d}^{*} \boldsymbol{T}=\boldsymbol{e}_{\mu} T^{\mu \nu}{ }_{; \nu} \sqrt{|g|} \boldsymbol{\omega}^{0} \wedge \boldsymbol{\omega}^{1} \wedge \boldsymbol{\omega}^{2} \wedge \boldsymbol{\omega}^{3}dT=eμTμν;ν|g|ω0ω1ω2ω3

BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY

This chapter is entirely Track 2. As preparation, one needs to have covered (1) Chapter 4 (differential forms) and (2) Chapter 14 (computation of curvature).
In reading it, one will be helped by Chapters 9-11 and 13.
It is not needed as preparation for any later chapter, but it will be helpful in Chapter 17 (Einstein field equations).

§15.1. BIANCHI IDENTITIES IN BRIEF

Geometry gives instructions to matter, but how does matter manage to give instructions to geometry? Geometry conveys its instructions to matter by a simple handle: "pursue a world line of extremal lapse of proper time (geodesic)." What is the handle by which matter can act back on geometry? How can one identify the right handle when the metric geometry of Riemann and Einstein has scores of interesting features? Physics tells one what to look for: a machinery of coupling between gravitation (spacetime curvature) and source (matter; stress-energy tensor T T T\boldsymbol{T}T ) that will guarantee the automatic conservation of the source ( T = 0 ) ( T = 0 ) (grad*T=0)(\boldsymbol{\nabla} \cdot \boldsymbol{T}=0)(T=0). Physics therefore asks mathematics: "What tensor-like feature of the geometry is automatically conserved?" Mathematics comes back with the answer: "The Einstein tensor." Physics queries, "How does this conservation come about?" Mathematics, in the person of Élie Cartan, replies, "Through the principle that 'the boundary of a boundary is zero'" (Box 15.1).
Actually, two features of the curvature are automatically conserved; or, otherwise stated, the curvature satisfies two Bianchi identities, the subject of this chapter. Both features of the curvature, both "geometric objects," lend themselves to representation in diagrams, moreover, diagrams that show in action the principle that "the boundary of a boundary is zero." In this respect, the geometry of spacetime shows a striking analogy to the field of Maxwell electrodynamics.
In electrodynamics there are four potentials that are united in the 1 -form A A A-=\boldsymbol{A} \equivA A μ d x μ A μ d x μ A_(mu)dx^(mu)A_{\mu} \boldsymbol{d} x^{\mu}Aμdxμ. Out of this quantity by differentiation follows the Faraday, F = d A F = d A F=dA\boldsymbol{F}=\boldsymbol{d} \boldsymbol{A}F=dA. This
Identities and conservation of the source: electromagnetism and gravitation compared:
field satisfies the identity d F = 0 d F = 0 dF=0\boldsymbol{d F}=0dF=0 (identity, yes; identity lending itself to the definition
d F 0 d F 0 dF-=0\boldsymbol{d} \boldsymbol{F} \equiv 0dF0 of a conserved source, no).
In gravitation there are ten potentials (metric coefficients g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν ) that are united in the metric tensor g = g μ ν d x μ d x ν g = g μ ν d x μ d x ν g=g_(mu nu)dx^(mu)ox dx^(nu)\boldsymbol{g}=g_{\mu \nu} \boldsymbol{d} x^{\mu} \otimes \boldsymbol{d} x^{\nu}g=gμνdxμdxν. Out of this quantity by two differentiations follows the curvature operator
= 1 4 e μ e ν R α β μ ν d x α d x β . = 1 4 e μ e ν R α β μ ν d x α d x β . ℜ=(1)/(4)e_(mu)^^e_(nu)R_(alpha beta)^(mu nu)dx^(alpha)^^dx^(beta).\Re=\frac{1}{4} \boldsymbol{e}_{\mu} \wedge \boldsymbol{e}_{\nu} R_{\alpha \beta}^{\mu \nu} \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} .=14eμeνRαβμνdxαdxβ.
This curvature operator satisfies the Bianchi identity d R = 0 d R = 0 dR=0\boldsymbol{d} \mathscr{R}=0dR=0, where now " d d d\boldsymbol{d}d " is a generalization of Cartan's exterior derivative, described more fully in Chapter 14 (again an identity, but again one that does not lend itself to the definition of a conserved source).
In electromagnetism, one has to go to the dual, F F ^(**)F{ }^{\boldsymbol{*}} \boldsymbol{F}F, to have any feature of the field that offers a handle to the source, d F = 4 π J d F = 4 π J d^(**F)=4pi^(**)J\boldsymbol{d}^{\boldsymbol{*} \boldsymbol{F}}=4 \pi^{*} \boldsymbol{J}dF=4πJ. The conservation of the source, d J = 0 d J = 0 d^(**)J=0\boldsymbol{d}^{*} \boldsymbol{J}=0dJ=0, appears as a consequence of the identity d d F = 0 d d F = 0 dd^(**)F=0\boldsymbol{d} \boldsymbol{d}^{*} \boldsymbol{F}=0ddF=0; or, by a rewording of the reasoning (Box 15.1), as a consequence of the vanishing of the boundary of a boundary.
(continued on page 370 )
d d F 0 d d F 0 dd^(**)F-=0\boldsymbol{d} \boldsymbol{d}^{*} \boldsymbol{F} \equiv 0ddF0 plus Maxwell equations d J = 0 d J = 0 Longrightarrowd^(**)J=0\Longrightarrow \boldsymbol{d}^{*} \boldsymbol{J}=0dJ=0

Box 15.1 THE BOUNDARY OF A BOUNDARY IS ZERO

A. The Idea in Its 1-2-3-Dimensional Form

Begin with an oriented cube or approximation to a cube (3-dimensional).
Its boundary is composed of six oriented faces, each two-dimensional. Orientation of each face is indicated by an arrow.
Boundary of any one oriented face consists of four oriented edges or arrows, each one-dimensional.
Every edge unites one face with another. No edge stands by itself in isolation.
"Sum" over all these edges, with due regard to sign. Find that any given edge is counted twice, once going one way, once going the other.
Conclude that the one-dimensional boundary of the two-dimensional boundary of the three-dimensional cube is identically zero.

Box 15.1 (continued)

B. The Idea in Its 2-3-4-Dimensional Form

Begin with an oriented four-dimensional cube or approximation thereto. The coordinates of the typical corner of the four-cube may be taken to be ( t 0 ± 1 2 Δ t , x 0 ± 1 2 Δ x , y 0 ± 1 2 Δ y , z 0 ± 1 2 Δ z ) t 0 ± 1 2 Δ t , x 0 ± 1 2 Δ x , y 0 ± 1 2 Δ y , z 0 ± 1 2 Δ z {:t_(0)+-(1)/(2)Delta t,x_(0)+-(1)/(2)Delta x,y_(0)+-(1)/(2)Delta y,z_(0)+-(1)/(2)Delta z)\left.t_{0} \pm \frac{1}{2} \Delta t, x_{0} \pm \frac{1}{2} \Delta x, y_{0} \pm \frac{1}{2} \Delta y, z_{0} \pm \frac{1}{2} \Delta z\right)t0±12Δt,x0±12Δx,y0±12Δy,z0±12Δz); and, accordingly, a sample corner itself, in an obvious abbreviation, is conveniently abbreviated +--+ . There are 16 of these corners. Less complicated in appearance than the 4 -cube itself are
its three-dimensional faces, which are "exploded off of it" into the surrounding area of the diagram, where they can be inspected in detail.
The boundary of the 4 -cube is composed of eight oriented hyperfaces, each of them three-dimensional (top hyperface with extension Δ x Δ y Δ z Δ x Δ y Δ z Delta x Delta y Delta z\Delta x \Delta y \Delta zΔxΔyΔz, for example; a "front" hyperface with extension Δ t Δ y Δ z ; Δ t Δ y Δ z ; Delta t Delta y Delta z;\Delta t \Delta y \Delta z ;ΔtΔyΔz; etc.)

x z y x z y sum_(x)^(z)y\sum_{x}^{z} yxzy
Boundary of any one hyperface ("cube") consists of six oriented faces, each two-dimensional.
Every face (for example, the hatched face Δ x Δ y Δ x Δ y Delta x Delta y\Delta x \Delta yΔxΔy in the lower lefthand corner) unites one hypersurface with another (the " 3 -cube side face" Δ t Δ x Δ y Δ t Δ x Δ y Delta t Delta x Delta y\Delta t \Delta x \Delta yΔtΔxΔy in the lower lefthand corner with the " 3 -cube top face" Δ x Δ y Δ z Δ x Δ y Δ z Delta x Delta y Delta z\Delta x \Delta y \Delta zΔxΔyΔz, in this example). No face stands by itself in isolation. The three-dimensional boundary of the 4 -cube exposes no 2 -surface to the outside world. It is faceless.
"Sum" over all these faces, with due regard to orientation. Find any given face is counted twice, once with one orientation, once with the opposite orientation.
Conclude that the two-dimensional boundary of the three-dimensional boundary of the four-dimensional cube is identically zero.

C. The Idea in Its General Abstract Form

= 0 = 0 del del=0\partial \partial=0=0 (the boundary of a boundary is zero).

D. Idea Behind Application to Gravitation and Electromagnetism

The one central point is a law of conservation (conservation of charge; conservation of momentum-energy).
The other central point is "automatic fulfillment" of this conservation law.
"Automatic conservation" requires that source not be an agent free to vary arbitrarily from place to place and instant to instant.
Source needs a tie to something that, while having degrees of freedom of its own, will cut down the otherwise arbitrary degrees of freedom of the source sufficiently to guarantee that the source automatically fulfills the conservation law. Give the name "field" to this something.
Define this field and "wire it up" to the source in such a way that the conservation of the source shall be an automatic consequence of the "zero boundary of a boundary." Or, more explicitly: Conservation demands no creation or destruction of source inside the four-dimensional cube shown in the diagram. Equivalently, integral of "creation events" (integral of d J d J d^(**)J\boldsymbol{d}^{*} \boldsymbol{J}dJ for electric charge; integral of d T d T d^(**)T\boldsymbol{d}^{*} \boldsymbol{T}dT for energy-momentum) over this four-dimensional region is required to be zero.
Integral of creation over this four-dimensional region translates into integral of source density-current ( J J (^(**)J:}\left({ }^{*} \boldsymbol{J}\right.(J or T ) T {:^(**)T)\left.{ }^{*} \boldsymbol{T}\right)T) over three-dimensional boundary of this region. This boundary consists of eight hyperfaces, each taken with due regard to orientation. Integral over upper hyperface (" Δ x Δ y Δ z Δ x Δ y Δ z Delta x Delta y Delta z\Delta x \Delta y \Delta zΔxΔyΔz )" gives amount of source present at later moment; over lower hyperface gives amount of souce present at earlier moment; over such hyperfaces as " Δ t Δ x Δ y Δ t Δ x Δ y Delta t Delta x Delta y\Delta t \Delta x \Delta yΔtΔxΔy " gives outflow of source over intervening period of time. Conservation demands that sum of these eight three-dimensional integrals shall be zero (details in Chapter 5).

Box 15.1 (continued)

Vanishing of this sum of three-dimensional integrals states the conservation requirement, but does not provide the machinery for "automatically" (or, in mathematical terms, "identically") meeting this requirement. For that, turn to principle that "boundary of a boundary is zero."
Demand that integral of source density-current over any oriented hyperface V V V\mathscr{V}V (three-dimensional region; "cube") shall equal integral of field over faces of this "cube" (each face being taken with the appropriate orientation and the cube being infinitesimal):
4 π V J = V F ; 8 π V T = V ( moment of rotation ) . 4 π V J = V F ; 8 π V T = V (  moment of   rotation  ) . 4piint_(V)**J=int_(del V)^(**)F;quad8piint_(V)**T=int_(del V)((" moment of ")/(" rotation ")).4 \pi \int_{V} * \boldsymbol{J}=\int_{\partial V}{ }^{*} \boldsymbol{F} ; \quad 8 \pi \int_{V} * \boldsymbol{T}=\int_{\partial V}\binom{\text { moment of }}{\text { rotation }} .4πVJ=VF;8πVT=V( moment of  rotation ).
Sum over the six faces of this cube and continue summing until the faces of all eight cubes are covered. Find that any given face (as, for example, the hatched face in the diagram) is counted twice, once with one orientation, once with the other ("boundary of a boundary is zero"). Thus is guaranteed the conservation of source: integral of source density-current over three-dimensional boundary of four-dimensional region is automatically zero, making integral of creation over interior of that four-dimensional region also identically zero.
Repeat calculation with boundary of that four-dimensional region slightly displaced in one locality [the "bubble differentiation" of Tomonaga (1946) and Schwinger (1948)], and conclude that conservation is guaranteed, not only in the four-dimensional region as a whole, but at every point within it, and, by extension, everywhere in spacetime.

E. Relation of Source to Field

One view: Source is primary. Field may have other duties, but its prime duty is to serve as "slave" of source. Conservation of source comes first; field has to adjust itself accordingly.
Alternative view: Field is primary. Field takes the responsibility of seeing to it that the source obeys the conservation law. Source would not know what to do in absence of the field, and would not even exist. Source is "built" from field. Conservation of source is consequence of this construction.
One model illustrating this view in an elementary context: Concept of "classical" electric charge as nothing but "electric lines of force trapped in the topology of a multiply connected space" [Weyl (1924b); Wheeler (1955); Misner and Wheeler (1957)].
On any view: Integral of source density-current over any three-dimensional region (a "cube" in simplified analysis above) equals integral of field over boundary of this region (the six faces of the cube above). No one has ever found any other way to understand the correlation between field law and conservation law.

F. Electromagnetism as a Model: How to "Wire Up" Source to Field to Give Automatic Conservation of Source Via " = 0 = 0 del del=0\partial \partial=0=0 " in Its 2-3-4-Dimensional Form

Conservation means zero creation of charge (zero creation in four-dimensional region Ω Ω Omega\OmegaΩ ).
Conservation therefore demands zero value for integral of charge density-current over three-dimensional boundary of this volume; thus,
0 = Ω J μ x μ d 4 Ω = Ω J μ d 3 Σ μ 0 = Ω J μ x μ d 4 Ω = Ω J μ d 3 Σ μ 0=int_(Omega)(delJ^(mu))/(delx^(mu))d^(4)Omega=int_(del Omega)J^(mu)d^(3)Sigma_(mu)0=\int_{\Omega} \frac{\partial J^{\mu}}{\partial x^{\mu}} d^{4} \Omega=\int_{\partial \Omega} J^{\mu} d^{3} \Sigma_{\mu}0=ΩJμxμd4Ω=ΩJμd3Σμ
in the Track-1 language of Chapters 3 and 5. Equivalently, in the coordinate-free abstract language of § § 4.3 4.6 § § 4.3 4.6 §§4.3-4.6\S \S 4.3-4.6§§4.34.6, one has
0 = Ω d J = Ω J , 0 = Ω d J = Ω J , 0=int_(Omega)d^(**)J=int_(del Omega)**J,0=\int_{\Omega} \boldsymbol{d}^{*} \boldsymbol{J}=\int_{\partial \Omega} * \boldsymbol{J},0=ΩdJ=ΩJ,
where
J = J 123 d x 1 d x 2 d x 3 + J 023 d x 0 d x 2 d x 3 + J 031 d x 0 d x 3 d x 1 + J 012 d x 0 d x 1 d x 2 J = J 123 d x 1 d x 2 d x 3 + J 023 d x 0 d x 2 d x 3 + J 031 d x 0 d x 3 d x 1 + J 012 d x 0 d x 1 d x 2 {:[**J=^(**)J_(123)dx^(1)^^dx^(2)^^dx^(3)+^(**)J_(023)dx^(0)^^dx^(2)^^dx^(3)],[+^(**)J_(031)dx^(0)^^dx^(3)^^dx^(1)+^(**)J_(012)dx^(0)^^dx^(1)^^dx^(2)]:}\begin{aligned} * \boldsymbol{J}= & { }^{*} J_{123} \boldsymbol{d} x^{1} \wedge \boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3}+{ }^{*} J_{023} \boldsymbol{d} x^{0} \wedge \boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3} \\ & +{ }^{*} J_{031} \boldsymbol{d} x^{0} \wedge \boldsymbol{d} x^{3} \wedge \boldsymbol{d} x^{1}+{ }^{*} J_{012} \boldsymbol{d} x^{0} \wedge \boldsymbol{d} x^{1} \wedge \boldsymbol{d} x^{2} \end{aligned}J=J123dx1dx2dx3+J023dx0dx2dx3+J031dx0dx3dx1+J012dx0dx1dx2
("eggcrate-like structure" of the 3 -form of charge-density and current-density).
Fulfill this conservation requirement automatically ("identically") through the principle that "the boundary of a boundary is zero" by writing 4 π J = d F 4 π J = d F 4pi^(**)J=d^(**)F4 \pi^{*} \boldsymbol{J}=\boldsymbol{d}^{*} \boldsymbol{F}4πJ=dF; thus,
4 π Ω J = Ω d F = Ω ( zero 1 ) F 0 4 π Ω J = Ω d F = Ω (  zero  1 ) F 0 4piint_(del Omega)**J=int_(del Omega)d^(**)F=int_(del del Omega(" zero "1))**F-=04 \pi \int_{\partial \Omega} * \boldsymbol{J}=\int_{\partial \Omega} \boldsymbol{d}^{*} \boldsymbol{F}=\int_{\partial \partial \Omega(\text { zero } 1)} * \boldsymbol{F} \equiv 04πΩJ=ΩdF=Ω( zero 1)F0
or, in Track-1 language, write 4 π J μ = F μ ν ; ν 4 π J μ = F μ ν ; ν 4piJ^(mu)=F^(mu nu)_(;nu)4 \pi J^{\mu}=F^{\mu \nu}{ }_{; \nu}4πJμ=Fμν;ν, and have
4 π Ω J μ d 3 Σ μ = Ω F ; ν μ ν d 3 Σ μ = Ω ( z e r o ! ) F μ α d 2 Σ μ α 0 . 4 π Ω J μ d 3 Σ μ = Ω F ; ν μ ν d 3 Σ μ = Ω ( z e r o ! ) F μ α d 2 Σ μ α 0 . 4piint_(del Omega)J^(mu)d^(3)Sigma_(mu)=int_(del Omega)F_(;nu)^(mu nu)d^(3)Sigma_(mu)=int_(del del Omega(zero!))F^(mu alpha)d^(2)Sigma_(mu alpha)-=0.4 \pi \int_{\partial \Omega} J^{\mu} d^{3} \Sigma_{\mu}=\int_{\partial \Omega} F_{; \nu}^{\mu \nu} d^{3} \Sigma_{\mu}=\int_{\partial \partial \Omega(z e r o!)} F^{\mu \alpha} d^{2} \Sigma_{\mu \alpha} \equiv 0 .4πΩJμd3Σμ=ΩF;νμνd3Σμ=Ω(zero!)Fμαd2Σμα0.
In other words, half of Maxwell's equations in their familiar flat-space form,
div E = E = 4 π ρ , curl B = × B = E ˙ + 4 π J , div E = E = 4 π ρ , curl B = × B = E ˙ + 4 π J , div E=grad*E=4pi rho,quad curl B=grad xx B=E^(˙)+4pi J,\operatorname{div} \boldsymbol{E}=\boldsymbol{\nabla} \cdot \boldsymbol{E}=4 \pi \rho, \quad \operatorname{curl} \boldsymbol{B}=\boldsymbol{\nabla} \times \boldsymbol{B}=\dot{\boldsymbol{E}}+4 \pi \boldsymbol{J},divE=E=4πρ,curlB=×B=E˙+4πJ,
"wire up" the source to the field in such a way that the law of conservation of source follows directly from " Ω = 0 Ω = 0 del del Omega=0\partial \partial \Omega=0Ω=0."

G. Electromagnetism Also Employs " = 0 = 0 del del=0\partial \partial=0=0 " in its 1-2-3-Dimensional Form ('No Magnetic Charge'")

Magnetic charge is linked with field via 4 π J mag = d F 4 π J mag  = d F 4piJ_("mag ")=dF4 \pi \boldsymbol{J}_{\text {mag }}=\boldsymbol{d F}4πJmag =dF (see point F F F\mathbf{F}F above for translation of this compact Track-2 language into equivalent Track-1 terms). Absence of

Box 15.1 (continued)

any magnetic charge says that integral of J mag J mag  J_("mag ")\boldsymbol{J}_{\text {mag }}Jmag  over any 3 -volume V V V\mathfrak{V}V is necessarily zero; or ("integration by parts," generalized Stokes theorem)
0 = V d F = V F = ( total magnetic flux exiting through V ) . 0 = V d F = V F = (  total magnetic flux   exiting through  V ) . 0=int_(V)dF=int_(delV)F=((" total magnetic flux ")/(" exiting through "delV)).0=\int_{\mathscr{V}} \boldsymbol{d} \boldsymbol{F}=\int_{\partial \mathscr{V}} \boldsymbol{F}=\binom{\text { total magnetic flux }}{\text { exiting through } \partial \mathscr{V}} .0=VdF=VF=( total magnetic flux  exiting through V).
In order to satisfy this requirement "automatically," via principle that "the boundary of a boundary is zero," write F = d A F = d A F=dA\boldsymbol{F}=\boldsymbol{d} \boldsymbol{A}F=dA ("expression of field in terms of 4-potential"), and have
V F = V d A = V ( zero ) A 0 . V F = V d A = V (  zero  ) A 0 . int_(del V)F=int_(del V)dA=int_(del del V(" zero "))A-=0.\int_{\partial V} \boldsymbol{F}=\int_{\partial V} \boldsymbol{d} \boldsymbol{A}=\int_{\partial \partial V(\text { zero })} \boldsymbol{A} \equiv 0 .VF=VdA=V( zero )A0.

H. Structure of Electrodynamics in Outline Form

In gravitation physics, one has to go to the "double dual" (two pairs of alternating indices, two places to take the dual) G = G = G=**^(**)\boldsymbol{G}=\boldsymbol{*}^{\boldsymbol{*}}G= of Riemann to have a feature of the field that offers a handle to the source:
G = Tr G = Einstein = 8 π T = 8 π × ( density of energy-momentum ) . G = Tr G =  Einstein  = 8 π T = 8 π × (  density of energy-momentum  ) . G=Tr G=" Einstein "=8pi T=8pi xx(" density of energy-momentum ").\boldsymbol{G}=\operatorname{Tr} \boldsymbol{G}=\text { Einstein }=8 \pi \boldsymbol{T}=8 \pi \times(\text { density of energy-momentum }) .G=TrG= Einstein =8πT=8π×( density of energy-momentum ).
The conservation of the source T e μ T μ ν ω ν T e μ T μ ν ω ν T-=e_(mu)T^(mu)_(nu)omega^(nu)\boldsymbol{T} \equiv \boldsymbol{e}_{\mu} T^{\mu}{ }_{\nu} \boldsymbol{\omega}^{\nu}TeμTμνων can be stated T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0. But better suited for the present purpose is the form (see Chapter 14 and exercise 14.18)

I. Structure of Geometrodynamics in Outline Form

d T = 0 , d G 0 plus Einstein field d T = 0 , d G 0  plus Einstein field  d^(**)T=0,quadd^(**)G-=0" plus Einstein field "\boldsymbol{d}^{*} \boldsymbol{T}=0, \quad \boldsymbol{d}^{*} \boldsymbol{G} \equiv 0 \text { plus Einstein field }dT=0,dG0 plus Einstein field 
where
T e μ T ν μ ( ω ν ) = e μ T μ ν d 3 Σ ν . T e μ T ν μ ω ν = e μ T μ ν d 3 Σ ν . ^(**)T-=e_(mu)T_(nu)^(mu)(^(**)omega^(nu))=e_(mu)T^(mu nu)d^(3)Sigma_(nu).{ }^{*} \boldsymbol{T} \equiv \boldsymbol{e}_{\mu} T_{\nu}^{\mu}\left({ }^{*} \boldsymbol{\omega}^{\nu}\right)=\boldsymbol{e}_{\mu} T^{\mu \nu} d^{3} \Sigma_{\nu} .TeμTνμ(ων)=eμTμνd3Σν.
This conservation law arises as a consequence of the "contracted Bianchi identity", d G = 0 d G = 0 d^(**)G=0\boldsymbol{d}^{*} \boldsymbol{G}=0dG=0, again interpretable in terms of the vanishing of the boundary of a boundary.
Figure 15.1.
Combine rotations associated with each of the six faces of the illustrated 3 -volume and end up with zero net rotation ("full Bianchi identity"). Reason: Contribution of any face is measured by change in a test vector A A A\boldsymbol{A}A carried in parallel transport around the perimeter of that face. Combine contributions of all faces and end up with each edge traversed twice, once in one direction, once in the other direction [boundary (here one-dimensional) of boundary (two-dimensional) of indicated three-dimensional figure is zero]. Detail: The vector A A A\boldsymbol{A}A, residing at the indicated site, is transported parallel to itself over to the indicated face, then carried around the perimeter of that face by parallel transport, experiencing in the process a rotation measured by the spacetime curvature associated with that face, then transported parallel to itself back to the original site. To the lowest relevant order of small quantities one can write
(change in A ) = Δ y Δ z R ( e y , e z ) A  (change in  A ) = Δ y Δ z R e y , e z A " (change in "A)=-Delta y Delta zR(e_(y),e_(z))A\text { (change in } \boldsymbol{A})=-\Delta y \Delta z \mathscr{R}\left(\boldsymbol{e}_{y}, \boldsymbol{e}_{z}\right) \boldsymbol{A} (change in A)=ΔyΔzR(ey,ez)A
in operator notation; or in coordinate language,
δ A α = R α β y z ( at x + Δ x ) A β Δ y Δ z . δ A α = R α β y z (  at  x + Δ x ) A β Δ y Δ z . -deltaA^(alpha)=R^(alpha)_(beta yz)(" at "x+Delta x)A^(beta)Delta y Delta z.-\delta A^{\alpha}=R^{\alpha}{ }_{\beta y z}(\text { at } x+\Delta x) A^{\beta} \Delta y \Delta z .δAα=Rαβyz( at x+Δx)AβΔyΔz.

§15.2. BIANCHI IDENTITY d R = 0 d R = 0 dR=0d \mathscr{R}=0dR=0 AS A MANIFESTATION OF "BOUNDARY OF BOUNDARY = 0 = 0 =0=0=0 "

Such is the story of the two Bianchi identities in outline form; it is now appropriate
Bianchi identity, d R 0 d R 0 dR-=0\boldsymbol{d} \mathscr{R} \equiv 0dR0, interpreted in terms of parallel transport around the six faces of a cube.
to fill in the details. Figure 15.1 illustrates the full Bianchi identity, d R = 0 d R = 0 dR=0\boldsymbol{d} \mathscr{R}=0dR=0 (see exercise 14.17), saying in brief, "The sum of the curvature-induced rotations associated with the six faces of any elementary cube is zero." The change in a vector A A A\boldsymbol{A}A associated with transport around the perimeter of the indicated face evaluated to the lowest relevant order of small quantities is given by
(15.1) δ A α = R α β y z ( at x + Δ x ) A β Δ y Δ z . (15.1) δ A α = R α β y z (  at  x + Δ x ) A β Δ y Δ z . {:(15.1)-deltaA^(alpha)=R^(alpha)_(beta yz)(" at "x+Delta x)A^(beta)Delta y Delta z.:}\begin{equation*} -\delta A^{\alpha}=R^{\alpha}{ }_{\beta y z}(\text { at } x+\Delta x) A^{\beta} \Delta y \Delta z . \tag{15.1} \end{equation*}(15.1)δAα=Rαβyz( at x+Δx)AβΔyΔz.
The opposite face gives a similar contribution, except that now the sign is reversed and the evaluation takes place at x x xxx rather than at x + Δ x x + Δ x x+Delta xx+\Delta xx+Δx. The combination of the contributions from the two faces gives
(15.2) R α β y z x A β Δ x Δ y Δ z , (15.2) R α β y z x A β Δ x Δ y Δ z , {:(15.2)(delR^(alpha)_(beta yz))/(del x)A^(beta)Delta x Delta y Delta z",":}\begin{equation*} \frac{\partial R^{\alpha}{ }_{\beta y z}}{\partial x} A^{\beta} \Delta x \Delta y \Delta z, \tag{15.2} \end{equation*}(15.2)RαβyzxAβΔxΔyΔz,
when Riemann normal coordinates are in use. In such coordinates, the vanishing of the total δ A α δ A α -deltaA^(alpha)-\delta A^{\alpha}δAα contributed by all six faces implies
(15.3) R α β y z ; x + R α β z z ; y + R β x y ; z α = 0 . (15.3) R α β y z ; x + R α β z z ; y + R β x y ; z α = 0 . {:(15.3)R^(alpha)_(beta yz;x)+R^(alpha)_(beta zz;y)+R_(beta xy;z)^(alpha)=0.:}\begin{equation*} R^{\alpha}{ }_{\beta y z ; x}+R^{\alpha}{ }_{\beta z z ; y}+R_{\beta x y ; z}^{\alpha}=0 . \tag{15.3} \end{equation*}(15.3)Rαβyz;x+Rαβzz;y+Rβxy;zα=0.
Here semicolons (covariant derivatives) can be and have been inserted instead of commas (ordinary derivatives), because the two are identical in the context of Riemann normal coordinates; and the covariant version (15.3) generalizes itself to arbitrary curvilinear coordinates. Turn from an x y z x y z xyzx y zxyz cube to a cube defined by any set of coordinate axes, and write Bianchi's identity in the form
(15.4) R α β [ λ μ ; ν ] = 0 (15.4) R α β [ λ μ ; ν ] = 0 {:(15.4)R^(alpha)_(beta[lambda mu;nu])=0:}\begin{equation*} R^{\alpha}{ }_{\beta[\lambda \mu ; \nu]}=0 \tag{15.4} \end{equation*}(15.4)Rαβ[λμ;ν]=0
(See exercise 14.17 for one reexpression of this identity in the abstract coordinateindependent form, d R = 0 d R = 0 dR=0\boldsymbol{d} \mathscr{R}=0dR=0, and § 15.3 § 15.3 §15.3\S 15.3§15.3 for another.) This identity occupies much the same place in gravitation physics as that occupied by the identity d F = d d A 0 d F = d d A 0 dF=ddA-=0\boldsymbol{d} \boldsymbol{F}=\boldsymbol{d} \boldsymbol{d} \boldsymbol{A} \equiv 0dF=ddA0 in electromagnetism:
(15.5) F [ λ μ , v ] = F [ λ μ ; v ] = 0 (15.5) F [ λ μ , v ] = F [ λ μ ; v ] = 0 {:(15.5)F_([lambda mu,v])=F_([lambda mu;v])=0:}\begin{equation*} F_{[\lambda \mu, v]}=F_{[\lambda \mu ; v]}=0 \tag{15.5} \end{equation*}(15.5)F[λμ,v]=F[λμ;v]=0

§15.3. MOMENT OF ROTATION: KEY TO CONTRACTED BIANCHI IDENTITY

The contracted Bianchi identity, the identity that offers a "handle to couple to the source," was shown by Élie Cartan to deal with "moments of rotation" [Cartan (1928); Wheeler (1964b); Misner and Wheeler (1972)]. Moments are familiar in elementary mechanics. A rigid body will not remain at rest unless all the forces acting on it sum to zero:
(15.6) i F ( i ) = 0 (15.6) i F ( i ) = 0 {:(15.6)sum_(i)F^((i))=0:}\begin{equation*} \sum_{i} F^{(i)}=0 \tag{15.6} \end{equation*}(15.6)iF(i)=0
Although necessary, this condition is not sufficient. The sum of the moments of these forces about some point P P P\mathscr{P}P must also be zero:
(15.7) i ( P ( i ) P ) F ( i ) = 0 (15.7) i P ( i ) P F ( i ) = 0 {:(15.7)sum_(i)(P^((i))-P)^^F^((i))=0:}\begin{equation*} \sum_{i}\left(\mathscr{P}^{(i)}-\mathscr{P}\right) \wedge F^{(i)}=0 \tag{15.7} \end{equation*}(15.7)i(P(i)P)F(i)=0
Exactly what point these moments are taken about happily does not matter, and this for a simple reason. The arbitrary point in the vector product (15.7) has for coefficient the quantity Σ i F ( i ) Σ i F ( i ) Sigma_(i)F^((i))\Sigma_{i} F^{(i)}ΣiF(i), which already has been required to vanish. The situation is similar in the elementary cube of Figure 15.1. Here the rotation associated with a given face is the analog of the force F ( i ) F ( i ) F^((i))\boldsymbol{F}^{(i)}F(i) in mechanics. That the sum of these rotations vanishes when extended over all six faces of the cube is the analog of the vanishing of the sum of the forces F ( i ) F ( i ) F^((i))\boldsymbol{F}^{(i)}F(i).
What is the analog for curvature of the moment of the force that one encounters in mechanics? It is the moment of the rotation associated with a given face of the
Net moment of rotation over all six faces of a cube:
(1) described
(2) equated to integral of source, T T int^(**)T\int{ }^{*} \boldsymbol{T}T, over interior of cube
cube. The value of any individual moment depends on the reference point P P P\mathscr{P}P. However, the sum of these moments taken over all six faces of the cube will have a value independent of the reference point P P P\mathscr{P}P, for the same reason as in mechanics. Therefore P P P\mathscr{P}P can be taken where one pleases, inside the elementary cube or outside it. Moreover, the cube may be viewed as a bit of a hypersurface sliced through spacetime. Therefore P P P\mathscr{P}P can as well be off the slice as on it. It is only required that all distances involved be short enough that one obtains the required precision by calculating the moments and the sum of moments in a local Riemann-normal coordinate system. One thus arrives at a P P P\mathscr{P}P-independent totalized moment of rotation (not necessarily zero; gravitation is not mechanics!) associated with the cube in question.
Now comes the magic of "the boundary of the boundary is zero." Identify this net moment of rotation of the cube, evaluated by summing individual moments of rotation associated with individual faces, with the integral of the source densitycurrent (energy-momentum tensor T T ^(**)T{ }^{*} \boldsymbol{T}T ) over the interior of the 3-cube. Make this identification not only for the one 3 -cube, but for all eight 3 -cubes (hyperfaces) that bound the four-dimensional cube in Box 15.1. Sum the integrated source densitycurrent T T ^(**)T{ }^{*} \boldsymbol{T}T not only for the one hyperface of the 4 -cube, but for all eight hyperfaces. Thus have
4-cube ( source creation d T ) = 3-boundary of this 4-cube ( source current- density, T ) = these eight bounding 3 cubbes ( net moment of rotation associated with speci- fied cube ) (15.8) = eight bonding 3-cubes six fares given 3ding gicube ( moment of rotation associated with specified face of specified cube ) . 4-cube   source   creation  d T =  3-boundary   of this 4-cube  (  source current-   density,  T ) =  these eight   bounding  3  cubbes   net moment of rotation   associated with speci-   fied cube  (15.8) =  eight   bonding   3-cubes   six fares   given 3ding   gicube   moment of rotation   associated with specified   face of specified cube  . {:[int_("4-cube ")({:[" source "],[" creation "],[d^(**)T]:})=int_({:[" 3-boundary "],[" of this 4-cube "]:})((" source current- ")/(" density, "^(**)T))],[=sum_({:[" these eight "],[" bounding "],[3-" cubbes "]:})([" net moment of rotation "],[" associated with speci- "],[" fied cube "])],[(15.8)=sum_({:[" eight "],[" bonding "],[" 3-cubes "]:})ubrace(ubrace)_({:[" six fares "],[" given 3ding "],[" gicube "]:})([" moment of rotation "],[" associated with specified "],[" face of specified cube "]).]:}\begin{align*} \int_{\text {4-cube }}\left(\begin{array}{l} \begin{array}{l} \text { source } \\ \text { creation } \\ \boldsymbol{d}^{*} \boldsymbol{T} \end{array} \end{array}\right) & =\int_{\substack{\text { 3-boundary } \\ \text { of this 4-cube }}}\binom{\text { source current- }}{\text { density, }{ }^{*} \boldsymbol{T}} \\ & =\sum_{\substack{\text { these eight } \\ \text { bounding } \\ 3-\text { cubbes }}}\left(\begin{array}{l} \text { net moment of rotation } \\ \text { associated with speci- } \\ \text { fied cube } \end{array}\right) \\ & =\sum_{\substack{\text { eight } \\ \text { bonding } \\ \text { 3-cubes }}} \underbrace{}_{\substack{\text { six fares } \\ \text { given 3ding } \\ \text { gicube }}}\left(\begin{array}{l} \text { moment of rotation } \\ \text { associated with specified } \\ \text { face of specified cube } \end{array}\right) . \tag{15.8} \end{align*}4-cube ( source  creation dT)= 3-boundary  of this 4-cube ( source current-  density, T)= these eight  bounding 3 cubbes ( net moment of rotation  associated with speci-  fied cube )(15.8)= eight  bonding  3-cubes  six fares  given 3ding  gicube ( moment of rotation  associated with specified  face of specified cube ).
(3) conserved
Let the moments of rotation, not only for the six faces of one cube, but for all the faces of all the cubes, be taken with respect to one and the same point P P P\mathscr{P}P. Recall (Box 15.1) that any given face joins two cubes or hyperfaces. It therefore appears twice in the count of faces, once with one orientation ("sense of circumnavigation in parallel transport to evaluate rotation") and once with the opposite orientation. Therefore the double sum vanishes identically (boundary of a boundary is zero!) This identity establishes existence of a new geometric object, a feature of the curvature, that is conserved, and therefore provides a handle to which to couple a source. The desired result has been achieved. Now to translate it into standard mathematics!

§15.4. CALCULATION OF THE MOMENT OF ROTATION

It remains to find the tensorial character and value of this conserved Cartan moment of rotation that appertains to any elementary 3-volume. The rotation associated with the front face Δ y Δ z e y e z Δ y Δ z e y e z Delta y Delta ze_(y)^^e_(z)\Delta y \Delta z \boldsymbol{e}_{y} \wedge \boldsymbol{e}_{z}ΔyΔzeyez of the cube in Figure 15.1 will be represented by the bivector
(15.9) ( rotation associated with front Δ y Δ z face ) = e λ e μ R | λ μ | y z Δ y Δ z (15.9) (  rotation associated   with front  Δ y Δ z  face  ) = e λ e μ R | λ μ | y z Δ y Δ z {:(15.9)((" rotation associated ")/(" with front "Delta y Delta z" face "))=e_(lambda)^^e_(mu)R^(|lambda mu|)_(yz)Delta y Delta z:}\begin{equation*} \binom{\text { rotation associated }}{\text { with front } \Delta y \Delta z \text { face }}=\boldsymbol{e}_{\lambda} \wedge \boldsymbol{e}_{\mu} R^{|\lambda \mu|}{ }_{y z} \Delta y \Delta z \tag{15.9} \end{equation*}(15.9)( rotation associated  with front ΔyΔz face )=eλeμR|λμ|yzΔyΔz
located at P front = ( t 1 2 Δ t , x + Δ x , y + 1 2 Δ y , z + 1 2 Δ z ) P front  = t 1 2 Δ t , x + Δ x , y + 1 2 Δ y , z + 1 2 Δ z P_("front ")=(t-(1)/(2)Delta t,x+Delta x,y+(1)/(2)Delta y,z+(1)/(2)Delta z)\mathscr{P}_{\text {front }}=\left(t-\frac{1}{2} \Delta t, x+\Delta x, y+\frac{1}{2} \Delta y, z+\frac{1}{2} \Delta z\right)Pfront =(t12Δt,x+Δx,y+12Δy,z+12Δz). This equation uses Riemann normal coordinates; indices enclosed by strokes, as in | λ μ | | λ μ | |lambda mu||\lambda \mu||λμ|, are summed with the restriction λ < μ λ < μ lambda < mu\lambda<\muλ<μ. The moment of this rotation with respect to the point P P P\mathscr{P}P will be represented by the trivector
(15.10) ( moment of rotation associated with front Δ y Δ z face ) = ( P center of front face P ) e λ e μ R | λ μ | y z Δ y Δ z (15.10)  moment of rotation   associated with   front  Δ y Δ z  face  = P center  of front   face  P e λ e μ R | λ μ | y z Δ y Δ z {:(15.10){:([" moment of rotation "],[" associated with "],[" front "Delta y Delta z" face "])=_({:(P_("center ")^("of front "):}" face ":})-P)^^e_(lambda)^^e_(mu)R^(|lambda mu|)_(yz)Delta y Delta z:}\left.\left(\begin{array}{l} \text { moment of rotation } \tag{15.10}\\ \text { associated with } \\ \text { front } \Delta y \Delta z \text { face } \end{array}\right)=\underset{\substack{\left(\mathscr{P}_{\text {center }}^{\text {of front }} \\ \right. \text { face }}}{ }-\mathscr{P}\right) \wedge \boldsymbol{e}_{\lambda} \wedge \boldsymbol{e}_{\mu} R^{|\lambda \mu|}{ }_{y z} \Delta y \Delta z(15.10)( moment of rotation  associated with  front ΔyΔz face )=(Pcenter of front  face P)eλeμR|λμ|yzΔyΔz
Here neither P center front P center front  P_("center front ")\mathscr{P}_{\text {center front }}Pcenter front  nor P P P\mathscr{P}P has any well-defined meaning whatsoever as a vector, but their difference is a vector in the limit of infinitesimal separation, Δ P = Δ P = DeltaP=\Delta \mathscr{P}=ΔP= P center front P P center front  P P_("center front ")-P\mathscr{P}_{\text {center front }}-\mathscr{P}Pcenter front P. With the back face a similar moment of rotation is associated, with the opposite sign, and with P center front P center front  P_("center front ")\mathscr{P}_{\text {center front }}Pcenter front  replaced by P center back. P center back.  P_("center back. ")\mathscr{P}_{\text {center back. }}Pcenter back. . In the difference between the two terms, the factor P P P\mathscr{P}P is of no interest, because one is already assured it will cancel out [Bianchi identity (15.4); analog of Σ F ( i ) = 0 Σ F ( i ) = 0 SigmaF^((i))=0\Sigma \boldsymbol{F}^{(i)}=0ΣF(i)=0 in mechanics]. The difference P center front P center back P center front  P center back  P_("center front ")-P_("center back ")\mathscr{P}_{\text {center front }}-\mathscr{P}_{\text {center back }}Pcenter front Pcenter back  has the value Δ x e x Δ x e x Delta xe_(x)\Delta x \boldsymbol{e}_{x}Δxex. Summing over all six faces, one has
\left.\begin{array}{rl} \left(\begin{array}{l} \text { net moment of } \\ \text { rotation associated } \\ \text { with cube or hyper- } \\ \text { face } \Delta x \Delta y \Delta z \end{array}\right. \end{array}\right)=\left.\begin{aligned} & \\ & \\ & \begin{aligned} \boldsymbol{e}_{x} & \wedge \boldsymbol{e}_{\lambda} \wedge \boldsymbol{e}_{\mu} R^{|\lambda \mu|}{ }_{y z} \Delta x \Delta y \Delta z \text { (front and back) } \\ & +\boldsymbol{e}_{y} \end{aligned} \wedge \boldsymbol{e}_{\lambda} \wedge \boldsymbol{e}_{\mu} R^{\mid \lambda \mu}{ }_{z x} \Delta y \Delta z \Delta x \text { (sides) } \\ &+\boldsymbol{e}_{z} \tag{15.11} \end{aligned} \wedge \boldsymbol{e}_{\lambda} \wedge \boldsymbol{e}_{\mu} R^{\mid \lambda \mu}\right|_{x y} \Delta z \Delta x \Delta y \text { (top and bottom). } .\tag not allowed in aligned environment
This sum one recognizes as the value (on the volume element e x e y e z Δ x e x e y e z Δ x e_(x)^^e_(y)^^e_(z)Delta x\boldsymbol{e}_{x} \wedge \boldsymbol{e}_{y} \wedge \boldsymbol{e}_{z} \Delta xexeyezΔx Δ y Δ z Δ y Δ z Delta y Delta z\Delta y \Delta zΔyΔz ) of the 3-form
e ν e λ e μ R | λ μ | | | α β | d x ν d x α d x β e ν e λ e μ R | λ μ | | α β | d x ν d x α d x β e_(nu)^^e_(lambda)^^e_(mu)R^(|lambda mu|)|_(|alpha beta|)dx^(nu)^^dx^(alpha)^^dx^(beta)\left.\boldsymbol{e}_{\nu} \wedge \boldsymbol{e}_{\lambda} \wedge \boldsymbol{e}_{\mu} R^{|\lambda \mu|}\right|_{|\alpha \beta|} \boldsymbol{d} x^{\nu} \wedge \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta}eνeλeμR|λμ|||αβ|dxνdxαdxβ
Moreover this 3 -form is defined, and precisely defined, at a point, whereas (15.11), applying as it does to an extended region, does not lend itself to an analysis that is at the same time brief and precise. Therefore forego (15.11) in favor of the 3-form. Only remember, when it comes down to interpretation, that this 3 -form is to be
evaluated for the "cube" e x e y e z Δ x Δ y Δ z e x e y e z Δ x Δ y Δ z e_(x)^^e_(y)^^e_(z)Delta x Delta y Delta z\boldsymbol{e}_{x} \wedge \boldsymbol{e}_{y} \wedge \boldsymbol{e}_{z} \Delta x \Delta y \Delta zexeyezΔxΔyΔz. Now note that the "trivectorvalued moment-of-rotation 3 -form" can also be written as
(15.12) ( moment of rotation ) = d P R = e ν e λ e μ R | λ μ | | α β | d x ν d x α d x β . . . . . (15.12)  moment of   rotation  = d P R = e ν e λ e μ R | λ μ | | α β | d x ν d x α d x β . . . . . {:(15.12)({:[" moment of "],[" rotation "])=dP^^R=e_(nu)^^e_(lambda)^^e_(mu)R^(|lambda mu|)_(|alpha beta|)dx^(nu)^^dx^(alpha)^^dx^(beta).....:}:}\left(\begin{array}{l} \left.\begin{array}{l} \text { moment of } \\ \text { rotation } \end{array}\right)=\boldsymbol{d} \mathscr{P} \wedge \mathscr{R}=\boldsymbol{e}_{\nu} \wedge \boldsymbol{e}_{\lambda} \wedge \boldsymbol{e}_{\mu} R^{|\lambda \mu|}{ }_{|\alpha \beta|} \boldsymbol{d} x^{\nu} \wedge \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} . . . . . \tag{15.12} \end{array}\right.(15.12)( moment of  rotation )=dPR=eνeλeμR|λμ||αβ|dxνdxαdxβ.....
Here
(15.13) d P = e σ d x σ (15.13) d P = e σ d x σ {:(15.13)dP=e_(sigma)dx^(sigma):}\begin{equation*} \boldsymbol{d} \mathscr{P}=\boldsymbol{e}_{\sigma} \boldsymbol{d} x^{\sigma} \tag{15.13} \end{equation*}(15.13)dP=eσdxσ
is Cartan's ( 1 1 ) ( 1 1 ) ((1)/(1))\binom{1}{1}(11) unit tensor. Also \Re is the curvature operator, treated as a bivectorvalued 2-form:
(15.14) R = e λ e μ R | λ μ | | α β | d x α d x β (15.14) R = e λ e μ R | λ μ | | α β | d x α d x β {:(15.14)R=e_(lambda)^^e_(mu)R^(|lambda mu|)_(|alpha beta|)dx^(alpha)^^dx^(beta):}\begin{equation*} \mathscr{R}=\boldsymbol{e}_{\lambda} \wedge \boldsymbol{e}_{\mu} R^{|\lambda \mu|}{ }_{|\alpha \beta|} \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} \tag{15.14} \end{equation*}(15.14)R=eλeμR|λμ||αβ|dxαdxβ
Using the language of components as in (15.11), or the abstract language introduced in (15.12), one finds oneself dealing with a trivector. A trivector can be left a trivector, as, in quite another context, an element of 3 -volume on a hypersurface in 4 -space can be left as a trivector. However, there it is more convenient to take the dual representation, and speak of the element of volume as a vector. Denote by ***\star a duality operation that acts only on contravariant vectors, trizectors, etc. (but not on forms). Then in a Lorentz frame one has ( e 1 e 2 e 3 ) = e 0 e 1 e 2 e 3 = e 0 ***(e_(1)^^e_(2)^^e_(3))=e_(0)\star\left(\boldsymbol{e}_{1} \wedge \boldsymbol{e}_{2} \wedge \boldsymbol{e}_{3}\right)=\boldsymbol{e}_{0}(e1e2e3)=e0; but ( d x 3 ) = d x 3 d x 3 = d x 3 ***(dx^(3))=dx^(3)\star\left(\boldsymbol{d} x^{3}\right)=\boldsymbol{d} x^{3}(dx3)=dx3. More generally,
(15.15) ( e ν e λ e μ ) = ε ν λ μ σ e σ (15.15) e ν e λ e μ = ε ν λ μ σ e σ {:(15.15)***(e_(nu)^^e_(lambda)^^e_(mu))=epsi_(nu lambda mu)^(sigma)e_(sigma):}\begin{equation*} \star\left(\boldsymbol{e}_{\nu} \wedge \boldsymbol{e}_{\lambda} \wedge \boldsymbol{e}_{\mu}\right)=\varepsilon_{\nu \lambda \mu}{ }^{\sigma} \boldsymbol{e}_{\sigma} \tag{15.15} \end{equation*}(15.15)(eνeλeμ)=ενλμσeσ
In this notation, the "vector-valued moment-of-rotation 3-form" is
( moment of rotation ) = ( d P R ) = e σ ε ν λ μ σ R | λ μ | | α β | d x ν d x α d x β = e σ ( R ) v σ | α β | d x ν d x α d x β , (  moment   of rotation  ) = ( d P R ) = e σ ε ν λ μ σ R | λ μ | | α β | d x ν d x α d x β = e σ R v σ | α β | d x ν d x α d x β , {:[((" moment ")/(" of rotation "))=***(dP^^R)=e_(sigma)epsi_(nu lambda mu)^(sigma)R^(|lambda mu|)_(|alpha beta|)dx^(nu)^^dx^(alpha)^^dx^(beta)],[=e_(sigma)(^(**)R)_(v)^(sigma)_(|alpha beta|)dx^(nu)^^dx^(alpha)^^dx^(beta)","]:}\begin{aligned} \binom{\text { moment }}{\text { of rotation }} & =\star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})=\boldsymbol{e}_{\sigma} \varepsilon_{\nu \lambda \mu}{ }^{\sigma} R^{|\lambda \mu|}{ }_{|\alpha \beta|} \boldsymbol{d} x^{\nu} \wedge \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} \\ & =\boldsymbol{e}_{\sigma}\left({ }^{*} R\right)_{v}{ }^{\sigma}{ }_{|\alpha \beta|} \boldsymbol{d} x^{\nu} \wedge \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta}, \end{aligned}( moment  of rotation )=(dPR)=eσενλμσR|λμ||αβ|dxνdxαdxβ=eσ(R)vσ|αβ|dxνdxαdxβ,
or, in one more step,
(15.16) ( moment of rotation ) = ( d P R ) = e σ ( R ) ν σ ν τ d 3 Σ τ (15.16) (  moment   of rotation  ) = ( d P R ) = e σ R ν σ ν τ d 3 Σ τ {:(15.16)((" moment ")/(" of rotation "))=***(dP^^R)=e_(sigma)(^(**)R^(**))_(nu)^(sigma nu tau)d^(3)Sigma_(tau):}\begin{equation*} \binom{\text { moment }}{\text { of rotation }}=\star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})=\boldsymbol{e}_{\sigma}\left({ }^{*} R^{*}\right)_{\nu}{ }^{\sigma \nu \tau} d^{3} \Sigma_{\tau} \tag{15.16} \end{equation*}(15.16)( moment  of rotation )=(dPR)=eσ(R)νσντd3Στ
Here d 3 Σ τ d 3 Σ τ d^(3)Sigma_(tau)d^{3} \Sigma_{\tau}d3Στ is a notation for basis 3 -forms, as in Box 5.4 ; thus,
(15.17) d x ν d x α d x β = ε ν α β τ d 3 Σ τ (15.17) d x ν d x α d x β = ε ν α β τ d 3 Σ τ {:(15.17)dx^(nu)^^dx^(alpha)^^dx^(beta)=epsi^(nu alpha beta tau)d^(3)Sigma_(tau):}\begin{equation*} \boldsymbol{d} x^{\nu} \wedge \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta}=\varepsilon^{\nu \alpha \beta \tau} d^{3} \Sigma_{\tau} \tag{15.17} \end{equation*}(15.17)dxνdxαdxβ=εναβτd3Στ
(In a local Lorentz frame, d x 1 d x 2 d x 3 = d 3 Σ 0 d x 1 d x 2 d x 3 = d 3 Σ 0 dx^(1)^^dx^(2)^^dx^(3)=d^(3)Sigma_(0)\boldsymbol{d} x^{1} \wedge \boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3}=d^{3} \Sigma_{0}dx1dx2dx3=d3Σ0.)
Nothing is more central to the analysis of curvature than the formula (15.16). It starts with an element of 3 -volume and ends up giving the moment of rotation in that 3-volume. The tensor that connects the starting volume with the final moment, the "contracted double-dual" of Riemann, is so important that it deserves and receives a name of its own, G G G-=\boldsymbol{G} \equivG Einstein; thus
(15.18) ( Einstein ) σ τ G σ τ = G ν σ ν τ = ( R ) ν σ ν τ . (15.18) (  Einstein  ) σ τ G σ τ = G ν σ ν τ = R ν σ ν τ . {:(15.18)(" Einstein ")^(sigma tau)-=G^(sigma tau)=G_(nu)^(sigma nu tau)=(^(**)R^(**))_(nu)^(sigma nu tau).:}\begin{equation*} (\text { Einstein })^{\sigma \tau} \equiv G^{\sigma \tau}=G_{\nu}{ }^{\sigma \nu \tau}=\left({ }^{*} R^{*}\right)_{\nu}{ }^{\sigma \nu \tau} . \tag{15.18} \end{equation*}(15.18)( Einstein )στGστ=Gνσντ=(R)νσντ.
This tensor received attention in § § 13.5 § § 13.5 §§13.5\S \S 13.5§§13.5 and 14.2 , and also in the examples at the
end of Chapter 14. In terms of Einstein, the connection between element of 3-volume and "vector-valued moment of rotation" is
(15.19) ( moment of rotation ) = ( d P R ) = e σ G σ τ d 3 Σ τ (15.19) (  moment   of rotation  ) = ( d P R ) = e σ G σ τ d 3 Σ τ {:(15.19)((" moment ")/(" of rotation "))=***(dP^^R)=e_(sigma)G^(sigma tau)d^(3)Sigma_(tau):}\begin{equation*} \binom{\text { moment }}{\text { of rotation }}=\star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})=\boldsymbol{e}_{\sigma} G^{\sigma \tau} d^{3} \Sigma_{\tau} \tag{15.19} \end{equation*}(15.19)( moment  of rotation )=(dPR)=eσGστd3Στ
The amount of "vector-valued moment of rotation" contained in the element of 3 -volume d 3 Σ μ d 3 Σ μ d^(3)Sigma_(mu)d^{3} \Sigma_{\mu}d3Σμ is identified by general relativity with the amount of energy-momentum contained in that 3 -volume. However, defer this identification for now. Concentrate instead on the conservation properties of this moment of rotation. See them once in the formulation of integral calculus, as a consequence of the principle " 0 . " " 0 . " "del del-=0."" \partial \partial \equiv 0 . ""0." See them then a second time, in differential formulation, as a consequence of " d d 0 d d 0 dd-=0\boldsymbol{d} \boldsymbol{d} \equiv 0dd0."

§15.5. CONSERVATION OF MOMENT OF ROTATION SEEN FROM "BOUNDARY OF A BOUNDARY IS ZERO"

The moment of rotation defines an automatically conserved quantity. In other words, the value of the moment of rotation for an elementary 3-volume Δ x Δ y Δ z Δ x Δ y Δ z Delta x Delta y Delta z\Delta x \Delta y \Delta zΔxΔyΔz after the lapse of a time Δ t Δ t Delta t\Delta tΔt is equal to the value of the moment of rotation for the same 3-volume at the beginning of that time, corrected by the inflow of moment of rotation over the six faces of the 3 -volume in that time interval (quantities proportional to Δ y Δ z Δ t Δ y Δ z Δ t Delta y Delta z Delta t\Delta y \Delta z \Delta tΔyΔzΔt, etc.) Now verify this conservation of moment of rotation in the language of "the boundary of a boundary." Follow the pattern of equation (15.8), but translate the words into formulas, item by item. Evaluate the amount of moment of rotation created in the elementary 4 -cube Ω Ω Omega\OmegaΩ, and find
"creation" ¯ definition ( "creation of moment of rotation" in the elementary 4-cube of spacetime Ω ) = Ω d G ; ¯  definition   "creation of moment of   rotation" in the elementary   4-cube of spacetime  Ω = Ω d G ; bar(uarr)_(" definition ")int([" "creation of moment of "],[" rotation" in the elementary "],[" 4-cube of spacetime "Omega])=int_(Omega)d^(**)G;\underset{\text { definition }}{\bar{\uparrow}} \int\left(\begin{array}{l}\text { "creation of moment of } \\ \text { rotation" in the elementary } \\ \text { 4-cube of spacetime } \Omega\end{array}\right)=\int_{\Omega} \boldsymbol{d}^{*} \boldsymbol{G} ;¯ definition ( "creation of moment of  rotation" in the elementary  4-cube of spacetime Ω)=ΩdG;
Conservation of net moment of rotation:
(1) derived from " = 0 = 0 del del=0\partial \partial=0=0 "
(15.20) = eight bounding 3 -cubes six faces bounding specified 3-cube ( moment of rotation face ( P R ) associated with specified face of specified cube ) 0 . (15.20) =  eight bounding  3 -cubes   six faces bounding   specified 3-cube   moment of rotation  face  ( P R )  associated with   specified face of   specified cube  0 . {:(15.20)=uarrsum_({:[" eight bounding "],[3"-cubes "]:})sum_({:[" six faces bounding "],[" specified 3-cube "]:})***([" moment of rotation "],[int_("face ")(P^^R)],[" associated with "],[{:[" specified face of "],[" specified cube "]:}])-=0.:}\underset{\uparrow}{=} \sum_{\substack{\text { eight bounding } \tag{15.20}\\ 3 \text {-cubes }}} \sum_{\substack{\text { six faces bounding } \\ \text { specified 3-cube }}} \star\left(\begin{array}{l} \text { moment of rotation } \\ \int_{\text {face }}(\mathscr{P} \wedge \mathscr{R}) \\ \text { associated with } \\ \begin{array}{l} \text { specified face of } \\ \text { specified cube } \end{array} \end{array}\right) \equiv 0 .(15.20)= eight bounding 3-cubes  six faces bounding  specified 3-cube ( moment of rotation face (PR) associated with  specified face of  specified cube )0.
Here step 1 is the theorem of Stokes. Step 2 is the identification established by (15.19) between the Einstein tensor and the moment of rotation. Step 3 breaks down the integral over the entire boundary Ω Ω del Omega\partial \OmegaΩ into integrals over the individual 3-cubes that constitute this boundary. Moreover, in all these integrals, the star ***\star is treated as a constant and taken outside the sign of integration. The reason for such treatment is simple: the duality operation ***\star involves only the metric, and the metric is locally constant throughout the infinitesimal 4 -cube over the boundary of which the integration extends. Step 4 uses the formula
(15.21) d ( P R ) = d P R + P d R = d P R (15.21) d ( P R ) = d P R + P d R = d P R {:(15.21)d(P^^R)=dP^^R+P^^dR=dP^^R:}\begin{equation*} \boldsymbol{d}(\mathscr{P} \wedge \mathscr{R})=\boldsymbol{d} \mathscr{P} \wedge \mathscr{R}+\mathscr{P} \wedge \boldsymbol{d} \mathscr{R}=\boldsymbol{d} \mathscr{P} \wedge \mathscr{R} \tag{15.21} \end{equation*}(15.21)d(PR)=dPR+PdR=dPR
and the theorem of Stokes to express each 3-cube integral as an integral of P R P R P^^R\mathscr{P} \wedge \mathscr{R}PR over the two-dimensional boundary of that cube. The culminating step is 5 . It has nothing to do with the integrand. It depends solely on the principle 0 0 del del-=0\partial \partial \equiv 00.
In brief, the conservation of moment of rotation follows from two circumstances. (1) The moment of rotation associated with any elementary 3 -cube is by definition a net value, obtained by adding the six moments of rotation associated with the six faces of that cube. (2) When one sums these net values for all eight 3 -cubes in (15.20), which are the boundary of the elementary 4 -cube Ω Ω Omega\OmegaΩ, one counts the contribution of a given 2 -face twice, once with one sign and once with the opposite sign. In virtue of the principle that "the boundary of a boundary is zero," the conservation of moment of rotation is thus an identity.

§15.6. CONSERVATION OF MOMENT OF ROTATION EXPRESSED IN DIFFERENTIAL FORM

(2) derived from "dd = 0 = 0 =0=0=0 "
Every conservation law stated in integral form lends itself to restatement in differential form, and conservation of moment of rotation is no exception. The calculation is brief. Evaluate the generalized exterior derivative of the moment of rotation in three steps, and find that it vanishes; thus:
d G = d [ ( d P R ) ] = [ d ( d P R ) ] = [ d 2 P R d P d R ] = 0 } step 1 step 2 step 3 d G = d [ ( d P R ) ] = [ d ( d P R ) ] = d 2 P R d P d R = 0  step  1  step  2  step  3 {:[d^(**)G,=d[***(dP^^R)]],[,=***[d(dP^^R)]],[,=***[d^(2)P^^R-dP^^dR]],[,=0]}{:[" step "1],[" step "2],[" step "3]:}\left.\begin{array}{rl} \boldsymbol{d}^{*} \boldsymbol{G} & =\boldsymbol{d}[\star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})] \\ & =\star[\boldsymbol{d}(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})] \\ & =\star\left[\boldsymbol{d}^{2} \mathscr{P} \wedge \mathscr{R}-\boldsymbol{d} \mathscr{P} \wedge \boldsymbol{d} \mathscr{R}\right] \\ & =0 \end{array}\right\} \begin{aligned} & \text { step } 1 \\ & \text { step } 2 \\ & \text { step } 3 \end{aligned}dG=d[(dPR)]=[d(dPR)]=[d2PRdPdR]=0} step 1 step 2 step 3
Step 1 uses the relation d = d d = d d***=***d\boldsymbol{d} \star=\star \boldsymbol{d}d=d. The star duality and the generalized exterior derivative commute because when d d d\boldsymbol{d}d is applied to a contravariant vector, it acts as a covariant derivative, and when ***\star is applied to a covariant vector or 1 -form, it is without effect. Step 2 applies the standard rule for the action of d d d\boldsymbol{d}d on a product of tensor-valued forms [see equation (14.13b)]. Step 3 deals with two terms. The first term vanishes because the first factor in it vanishes; thus, d 2 P = 0 d 2 P = 0 d^(2)P=0\boldsymbol{d}^{2} \mathscr{P}=0d2P=0 [Cartan's equation of structure; expresses the "vanishing torsion" of the covariant derivative; see equation (14.26)]. The second term also vanishes, in this case, because the second factor in it vanishes; thus, d R = 0 d R = 0 dR=0\boldsymbol{d} \mathscr{R}=0dR=0 (the full Bianchi identity). Thus briefly is conservation of moment of rotation established.
Box 15.2 THE SOURCE OF GRAVITATION AND THE MOMENT OF ROTATION: THE TWO KEY QUANTITIES AND THE MOST USEFUL MATHEMATICAL REPRESENTATIONS FOR THEM
Energy-momentum as source of gravitation (curvature of spacetime) Moment of rotation as automatically conserved feature of the geometry
Representation as a vector-valued 3-form, a coordinate-independent geometric object
Machine to tell how much energymomentum is contained in an elementary 3 -volume: T = e σ T σ τ d 3 Σ τ T = e σ T σ τ d 3 Σ τ ^(**)T=e_(sigma)T^(sigma tau)d^(3)Sigma_(tau){ }^{*} \boldsymbol{T}=\boldsymbol{e}_{\sigma} T^{\sigma \tau} d^{3} \Sigma_{\tau}T=eσTστd3Στ
("dual of stress-energy tensor")
Machine to tell how much energymomentum is contained in an elementary 3 -volume: ^(**)T=e_(sigma)T^(sigma tau)d^(3)Sigma_(tau) ("dual of stress-energy tensor")| Machine to tell how much energymomentum is contained in an elementary 3 -volume: ${ }^{*} \boldsymbol{T}=\boldsymbol{e}_{\sigma} T^{\sigma \tau} d^{3} \Sigma_{\tau}$ | | :--- | | ("dual of stress-energy tensor") |
Machine to tell how much net moment of rotation-expressed as a vector-is obtained by adding the six moments of rotation associated with the six faces of the elementary 3-cube:
( d P R ) = G = e σ G σ τ d 3 Σ τ ( d P R ) = G = e σ G σ τ d 3 Σ τ ***(dP^^R)=^(**)G=e_(sigma)G^(sigma tau)d^(3)Sigma_(tau)\star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})={ }^{*} \boldsymbol{G}=\boldsymbol{e}_{\sigma} G^{\sigma \tau} d^{3} \Sigma_{\tau}(dPR)=G=eσGστd3Στ
("dual of Einstein")
Machine to tell how much net moment of rotation-expressed as a vector-is obtained by adding the six moments of rotation associated with the six faces of the elementary 3-cube: ***(dP^^R)=^(**)G=e_(sigma)G^(sigma tau)d^(3)Sigma_(tau) ("dual of Einstein")| Machine to tell how much net moment of rotation-expressed as a vector-is obtained by adding the six moments of rotation associated with the six faces of the elementary 3-cube: | | :--- | | $\star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})={ }^{*} \boldsymbol{G}=\boldsymbol{e}_{\sigma} G^{\sigma \tau} d^{3} \Sigma_{\tau}$ | | ("dual of Einstein") |
Representation as a 0 2 0 2 _(0)^(2){ }_{0}^{2}02 )-tensor (also a coordinate independent geometric object) Stress-energy tensor itself: T = e σ T σ τ e τ T = e σ T σ τ e τ T=e_(sigma)T^(sigma tau)e_(tau)\boldsymbol{T}=\boldsymbol{e}_{\sigma} T^{\sigma \tau} \boldsymbol{e}_{\tau}T=eσTστeτ Einstein itself: G = e σ G σ τ e τ G = e σ G σ τ e τ G=e_(sigma)G^(sigma tau)e_(tau)\boldsymbol{G}=\boldsymbol{e}_{\sigma} G^{\sigma \tau} \boldsymbol{e}_{\tau}G=eσGστeτ
Energy-momentum as source of gravitation (curvature of spacetime) Moment of rotation as automatically conserved feature of the geometry Representation as a vector-valued 3-form, a coordinate-independent geometric object "Machine to tell how much energymomentum is contained in an elementary 3 -volume: ^(**)T=e_(sigma)T^(sigma tau)d^(3)Sigma_(tau) ("dual of stress-energy tensor")" "Machine to tell how much net moment of rotation-expressed as a vector-is obtained by adding the six moments of rotation associated with the six faces of the elementary 3-cube: ***(dP^^R)=^(**)G=e_(sigma)G^(sigma tau)d^(3)Sigma_(tau) ("dual of Einstein")" Representation as a _(0)^(2) )-tensor (also a coordinate independent geometric object) Stress-energy tensor itself: T=e_(sigma)T^(sigma tau)e_(tau) Einstein itself: G=e_(sigma)G^(sigma tau)e_(tau)| | Energy-momentum as source of gravitation (curvature of spacetime) | Moment of rotation as automatically conserved feature of the geometry | | :---: | :---: | :---: | | Representation as a vector-valued 3-form, a coordinate-independent geometric object | Machine to tell how much energymomentum is contained in an elementary 3 -volume: ${ }^{*} \boldsymbol{T}=\boldsymbol{e}_{\sigma} T^{\sigma \tau} d^{3} \Sigma_{\tau}$ <br> ("dual of stress-energy tensor") | Machine to tell how much net moment of rotation-expressed as a vector-is obtained by adding the six moments of rotation associated with the six faces of the elementary 3-cube: <br> $\star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})={ }^{*} \boldsymbol{G}=\boldsymbol{e}_{\sigma} G^{\sigma \tau} d^{3} \Sigma_{\tau}$ <br> ("dual of Einstein") | | Representation as a ${ }_{0}^{2}$ )-tensor (also a coordinate independent geometric object) | Stress-energy tensor itself: $\boldsymbol{T}=\boldsymbol{e}_{\sigma} T^{\sigma \tau} \boldsymbol{e}_{\tau}$ | Einstein itself: $\boldsymbol{G}=\boldsymbol{e}_{\sigma} G^{\sigma \tau} \boldsymbol{e}_{\tau}$ |
Representation in language of
components (values depend on
choice of coordinate system)
Representation in language of components (values depend on choice of coordinate system)| Representation in language of | | :--- | | components (values depend on | | choice of coordinate system) |
T σ τ T σ τ T^(sigma tau)T^{\sigma \tau}Tστ G σ τ G σ τ G^(sigma tau)G^{\sigma \tau}Gστ
"Representation in language of components (values depend on choice of coordinate system)" T^(sigma tau) G^(sigma tau)| Representation in language of <br> components (values depend on <br> choice of coordinate system) | $T^{\sigma \tau}$ | $G^{\sigma \tau}$ | | :--- | :--- | :--- |
Conservation law in language of
components
Conservation law in language of components| Conservation law in language of | | :--- | | components |
T σ τ i τ = 0 T σ τ i τ = 0 T^(sigma tau)_(i tau)=0T^{\sigma \tau}{ }_{i \tau}=0Tστiτ=0 G σ τ 0 G σ τ 0 G^(sigma tau)-=0G^{\sigma \tau} \equiv 0Gστ0
Conservation in abstract lan-
guage, for the ( 2 0 ) ( 2 0 ) ((2)/(0))\binom{2}{0}(20)-tensor
Conservation in abstract lan- guage, for the ((2)/(0))-tensor| Conservation in abstract lan- | | :--- | | guage, for the $\binom{2}{0}$-tensor |
T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 G 0 G 0 grad*G-=0\boldsymbol{\nabla} \cdot \boldsymbol{G} \equiv 0G0
"Conservation law in language of components" T^(sigma tau)_(i tau)=0 G^(sigma tau)-=0 "Conservation in abstract lan- guage, for the ((2)/(0))-tensor" grad*T=0 grad*G-=0| Conservation law in language of <br> components | $T^{\sigma \tau}{ }_{i \tau}=0$ | $G^{\sigma \tau} \equiv 0$ | | :--- | :---: | :---: | | Conservation in abstract lan- <br> guage, for the $\binom{2}{0}$-tensor | $\boldsymbol{\nabla} \cdot \boldsymbol{T}=0$ | $\boldsymbol{\nabla} \cdot \boldsymbol{G} \equiv 0$ |
Conservation in abstract language, as translated into exterior derivative of the dual tensor (vec-
d T = 0 d T = 0 d^(**)T=0\boldsymbol{d}^{*} \boldsymbol{T}=0dT=0
d G 0 or d ( d P R ) 0 d G 0  or  d ( d P R ) 0 {:[d^(**)G-=0" or "],[d***(dP^^R)-=0]:}\begin{aligned} \boldsymbol{d}^{*} \boldsymbol{G} & \equiv 0 \text { or } \\ \boldsymbol{d} \star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R}) & \equiv 0 \end{aligned}dG0 or d(dPR)0
tor-valued 3-form)
Ω G 0 or Ω ( d P R ) 0 or Ω ( P R ) 0 Ω G 0  or  Ω ( d P R ) 0  or  Ω ( P R ) 0 {:[int_(del Omega)^(**)G-=0" or "],[***int_(del Omega)(dP^^R)-=0" or "],[***int_(del del Omega)(P^^R)-=0]:}\begin{aligned} \int_{\partial \Omega}{ }^{*} \boldsymbol{G} & \equiv 0 \text { or } \\ \star \int_{\partial \Omega}(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R}) & \equiv 0 \text { or } \\ \star \int_{\partial \partial \Omega}(\mathscr{P} \wedge \mathscr{R}) & \equiv 0 \end{aligned}ΩG0 or Ω(dPR)0 or Ω(PR)0
Same conservation law expressed in integral form for an element of 4 -volume Ω Ω Omega\OmegaΩ

§15.7. FROM CONSERVATION OF MOMENT OF ROTATION TO EINSTEIN'S GEOMETRODYNAMICS: A PREVIEW

Mass, or mass-energy, is the source of gravitation. Mass-energy is one component of the energy-momentum 4-vector. Energy and momentum are conserved. The amount of energy-momentum in the element of 3 -volume d 3 Σ d 3 Σ d^(3)Sigmad^{3} \boldsymbol{\Sigma}d3Σ is
(15.22) T = e σ T σ τ d 3 Σ τ (15.22) T = e σ T σ τ d 3 Σ τ {:(15.22)**T=e_(sigma)T^(sigma tau)d^(3)Sigma_(tau):}\begin{equation*} * \boldsymbol{T}=\boldsymbol{e}_{\sigma} T^{\sigma \tau} d^{3} \Sigma_{\tau} \tag{15.22} \end{equation*}(15.22)T=eσTστd3Στ
(see Box 15.2). Conservation of energy-momentum for an elementary 4-cube Ω Ω Omega\OmegaΩ expresses itself in the form
(15.23) Ω T = 0 . (15.23) Ω T = 0 . {:(15.23)int_(del Omega)^(**)T=0.:}\begin{equation*} \int_{\partial \Omega}{ }^{*} \boldsymbol{T}=0 . \tag{15.23} \end{equation*}(15.23)ΩT=0.
Einstein field equation "derived" from demand that (conservation of net moment of rotation) =>\Rightarrow (conservation of source)
This conservation is not an accident. According to Einstein and Cartan, it is "automatic"; and automatic, moreover, as a consequence of exact equality between energy-momentum and an automatically conserved feature of the geometry. What is this feature? It is the moment of rotation, which satisfies the law of automatic conservation,
(15.24) Ω G = 0 . (15.24) Ω G = 0 . {:(15.24)int_(del Omega)^(**)G=0.:}\begin{equation*} \int_{\partial \Omega}{ }^{*} \boldsymbol{G}=0 . \tag{15.24} \end{equation*}(15.24)ΩG=0.
In other words, the conservation of momentum-energy is to be made geometric in character and automatic in action by the following prescription: Identify the stressenergy tensor (up to a factor 8 π 8 π 8pi8 \pi8π, or 8 π G / c 4 8 π G / c 4 8pi G//c^(4)8 \pi G / c^{4}8πG/c4, or other factor that depends on choice of units) with the moment of rotation; thus,
(15.25) ( d P R ) = G = 8 π T ; (15.25) ( d P R ) = G = 8 π T ; {:(15.25)***(dP^^R)=^(**)G=8pi^(**)T;:}\begin{equation*} \star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})={ }^{*} \boldsymbol{G}=8 \pi^{*} \boldsymbol{T} ; \tag{15.25} \end{equation*}(15.25)(dPR)=G=8πT;
or equivalently (still in the language of vector-valued 3 -forms)
(15.26) ( moment of rotation ) = ( d P R ) = e σ G σ τ d 3 Σ τ = 8 π e σ T σ τ d 3 Σ τ (15.26) (  moment of   rotation  ) = ( d P R ) = e σ G σ τ d 3 Σ τ = 8 π e σ T σ τ d 3 Σ τ {:(15.26)((" moment of ")/(" rotation "))=***(dP^^R)=e_(sigma)G^(sigma tau)d^(3)Sigma_(tau)=8pie_(sigma)T^(sigma tau)d^(3)Sigma_(tau):}\begin{equation*} \binom{\text { moment of }}{\text { rotation }}=\star(\boldsymbol{d} \mathscr{P} \wedge \mathscr{R})=\boldsymbol{e}_{\sigma} G^{\sigma \tau} d^{3} \Sigma_{\tau}=8 \pi \boldsymbol{e}_{\sigma} T^{\sigma \tau} d^{3} \Sigma_{\tau} \tag{15.26} \end{equation*}(15.26)( moment of  rotation )=(dPR)=eσGστd3Στ=8πeσTστd3Στ
or, in the language of tensors,
(15.27) G = e σ G σ τ e τ = 8 π e σ T σ τ e τ = 8 π T (15.27) G = e σ G σ τ e τ = 8 π e σ T σ τ e τ = 8 π T {:(15.27)G=e_(sigma)G^(sigma tau)e_(tau)=8pie_(sigma)T^(sigma tau)e_(tau)=8pi T:}\begin{equation*} \boldsymbol{G}=\boldsymbol{e}_{\sigma} G^{\boldsymbol{\sigma} \tau} \boldsymbol{e}_{\tau}=8 \pi \boldsymbol{e}_{\sigma} T^{\sigma \tau} \boldsymbol{e}_{\tau}=8 \pi \boldsymbol{T} \tag{15.27} \end{equation*}(15.27)G=eσGστeτ=8πeσTστeτ=8πT
or, in the language of components,
(15.28) G σ τ = 8 π T σ τ (15.28) G σ τ = 8 π T σ τ {:(15.28)G^(sigma tau)=8piT^(sigma tau):}\begin{equation*} G^{\sigma \tau}=8 \pi T^{\sigma \tau} \tag{15.28} \end{equation*}(15.28)Gστ=8πTστ
(Einstein's field equation; more detail, and more on the question of uniqueness, will be found in Chapter 17; see also Box 15.3). Thus simply is all of general relativity tied to the principle that the boundary of a boundary is zero. No one has ever discovered a more compelling foundation for the principle of conservation of momentum and energy. No one has ever seen more deeply into that action of matter on space, and space on matter, which one calls gravitation.
In summary, the Einstein theory realizes the conservation of energy-momentum as the identity, "the boundary of a boundary is zero."

EXERCISES

Exercise 15.1. THE BOUNDARY OF THE BOUNDARY OF A 4-SIMPLEX

In the analysis of the development in time of a geometry lacking all symmetry, when one is compelled to resort to a computer, one can, as one option, break up the 4 -geometry into simplexes [four-dimensional analog of two-dimensional triangle, three-dimensional tetrahedron; vertices of "central simplex" conveniently considered to be at ( t , x , y , z ) = ( 0 , 1 , 1 , 1 ) ( t , x , y , z ) = ( 0 , 1 , 1 , 1 ) (t,x,y,z)=(0,1,1,1)(t, x, y, z)=(0,1,1,1)(t,x,y,z)=(0,1,1,1), ( 0 , 1 , 1 , 1 ) , ( 0 , 1 , 1 , 1 ) , ( 0 , 1 , 1 , 1 ) , ( 5 1 / 2 , 0 , 0 , 0 ) ( 0 , 1 , 1 , 1 ) , ( 0 , 1 , 1 , 1 ) , ( 0 , 1 , 1 , 1 ) , 5 1 / 2 , 0 , 0 , 0 (0,1,-1,-1),(0,-1,1,-1),(0,-1,-1,1),(5^(1//2),0,0,0)(0,1,-1,-1),(0,-1,1,-1),(0,-1,-1,1),\left(5^{1 / 2}, 0,0,0\right)(0,1,1,1),(0,1,1,1),(0,1,1,1),(51/2,0,0,0), for example], sufficiently numerous, and each sufficiently small, that the geometry inside each can be idealized as flat (Lorentzian), with all the curvature concentrated at the join between simplices (see discussion of dynamics of geometry via Regge calculus in Chapter 42). Determine ("give a mathematical

Box 15.3 OTHER IDENTITIES SATISFIED BY THE CURVATURE

(1) The source of gravitation is energy-momentum.
(2) Energy-momentum is expressed by stressenergy tensor (or by its dual) as a vector-valued 3 -form ("energy-momentum per unit 3 -volume").
(3) This source is conserved (no creation in an elementary spacetime 4 -cube).
These principles form the background for the probe in this chapter of the Bianchi identities. That is why two otherwise most interesting identities [Allendoerfer and Weil (1943); Chern ( 1955 , 1962 ) ( 1955 , 1962 ) (1955,1962)(1955,1962)(1955,1962) ] are dropped from attention. One deals with the 4-form
(1) Π = 1 24 π 2 g α γ g β δ R α β R γ δ (1) Π = 1 24 π 2 g α γ g β δ R α β R γ δ {:(1)Pi=(1)/(24pi^(2))g^(alpha gamma)g^(beta delta)R_(alpha beta)^^R_(gamma delta):}\begin{equation*} \Pi=\frac{1}{24 \pi^{2}} g^{\alpha \gamma} g^{\beta \delta} \mathscr{R}_{\alpha \beta} \wedge \mathscr{R}_{\gamma \delta} \tag{1} \end{equation*}(1)Π=124π2gαγgβδRαβRγδ
and the other with the 4 -form
Γ = 1 8 π 2 | det g μ ν | 1 / 2 ( R 12 R 30 + 13 02 (2) + R 10 23 ) . Γ = 1 8 π 2 det g μ ν 1 / 2 R 12 R 30 + 13 02 (2) + R 10 23 . {:[Gamma=(1)/(8pi^(2)|detg_(mu nu)|^(1//2))(R_(12)^^R_(30):}+ℜ_(13)^^ℜ_(02)],[(2){:+R_(10)^^ℜ_(23)).]:}\begin{align*} \Gamma=\frac{1}{8 \pi^{2}\left|\operatorname{det} g_{\mu \nu}\right|^{1 / 2}}\left(\mathscr{R}_{12} \wedge \mathscr{R}_{30}\right. & +\Re_{13} \wedge \Re_{02} \\ & \left.+\mathscr{R}_{10} \wedge \Re_{23}\right) . \tag{2} \end{align*}Γ=18π2|detgμν|1/2(R12R30+1302(2)+R1023).
Both quantities are built from the tensorial "curvature 2 -forms"
(3) α γ = 1 2 R α γ β δ d x β d x δ . (3) α γ = 1 2 R α γ β δ d x β d x δ . {:(3)ℜ_(alpha gamma)=(1)/(2)R_(alpha gamma beta delta)dx^(beta)^^dx^(delta).:}\begin{equation*} \Re_{\alpha \gamma}=\frac{1}{2} R_{\alpha \gamma \beta \delta} \boldsymbol{d} x^{\beta} \wedge \boldsymbol{d} x^{\delta} . \tag{3} \end{equation*}(3)αγ=12Rαγβδdxβdxδ.
The four-dimensional integral of either quantity over a four-dimensional region Ω Ω Omega\OmegaΩ has a value that (1) is a scalar, (2) is not identically equal to zero,
(3) depends on the boundary of the region of spacetime over which the integral is extended, but (4) is independent of any changes made in the
spacetime geometry interior to that surface (provided that these changes neither abandon the continuity nor change the connectivity of the 4 -geometry in that region). Property (1) kills any possibility of identifying the integral, a scalar, with energy-momentum, a 4 -vector. Property (2) kills it for the purpose of a conservation law, because it implies a non-zero creation in Ω Ω Omega\OmegaΩ.
Also omitted here is the Bel-Robinson tensor (see exercise 15.2), built bilinearly out of the curvature tensor, and other tensors for which see, e.g., Synge (1962).
One or all of these quantities may be found someday to have important physical content.
The integral of the 4 -form Γ Γ Gamma\GammaΓ of equation (2) over the entire manifold gives a number, an integer, the so-called Euler-Poincaré characteristic of the manifold, whenever the integral and the integer are well-defined. This result is the four-dimensional generalization of the Gauss-Bonnet integral, widely known in the context of two-dimensional geometry:
( Riemannian scalar curvature invariant (value 2 / a 2 for a sphere of radius a ) ) g 1 / 2 d 2 x  Riemannian scalar curvature   invariant (value  2 / a 2  for a sphere of radius  a  )  g 1 / 2 d 2 x int([" Riemannian scalar curvature "],[" invariant (value "2//a^(2)],[" for a sphere of radius "a" ) "])g^(1//2)d^(2)x\int\left(\begin{array}{l} \text { Riemannian scalar curvature } \\ \text { invariant (value } 2 / a^{2} \\ \text { for a sphere of radius } a \text { ) } \end{array}\right) g^{1 / 2} d^{2} x( Riemannian scalar curvature  invariant (value 2/a2 for a sphere of radius a ) )g1/2d2x
This integral has the value 8 π 8 π 8pi8 \pi8π for any closed, oriented, two-dimensional manifold with the topology of a 2 -sphere, no matter how badly distorted; and the value 0 for any 2 -torus, again no matter how rippled and twisted; and other equally specific values for other topologies.
description of") the boundary (three-dimensional) of such a simplex. Take one piece of this boundary and determine its boundary (two-dimensional). For one piece of this two-dimensional boundary, verify that there is at exactly one other place, and no more, in the bookkeeping on the boundary of a boundary, another two-dimensional piece that cancels it ("facelessness" of the 3-boundary of the simplex).

Exercise 15.2. THE BEL-ROBINSON TENSOR [Bel (1958, 1959, 1962),

Robinson (1959b), Sejnowski (1973); see also Pirani (1957) and Lichnerowicz (1962)].
Define the Bel-Robinson tensor by
(15.29) T α β γ δ = R α ρ γ σ R β δ σ + R α ρ γ γ σ R β ρ δ σ . (15.29) T α β γ δ = R α ρ γ σ R β δ σ + R α ρ γ γ σ R β ρ δ σ . {:(15.29)T_(alpha beta gamma delta)=R_(alpha rho gamma sigma)R_(beta)^(@)_(delta)^(sigma)+^(**)R_(alpha rho gamma gamma sigma)^(**)R_(beta)^(rho)_(delta)^(sigma).:}\begin{equation*} T_{\alpha \beta \gamma \delta}=R_{\alpha \rho \gamma \sigma} R_{\beta}{ }^{\circ}{ }_{\delta}{ }^{\sigma}+{ }^{*} R_{\alpha \rho \gamma \gamma \sigma}{ }^{*} R_{\beta}{ }^{\rho}{ }_{\delta}{ }^{\sigma} . \tag{15.29} \end{equation*}(15.29)Tαβγδ=RαργσRβδσ+RαργγσRβρδσ.
Show that in empty spacetime this tensor can be rewritten as
(15.30a) T α β γ δ = R α ρ γ σ R β ρ δ σ + R α ρ δ σ R β ρ γ σ 1 8 g α β g γ δ R ρ σ λ μ R ρ σ λ μ . (15.30a) T α β γ δ = R α ρ γ σ R β ρ δ σ + R α ρ δ σ R β ρ γ σ 1 8 g α β g γ δ R ρ σ λ μ R ρ σ λ μ . {:(15.30a)T_(alpha beta gamma delta)=R_(alpha rho gamma sigma)R_(beta)^(rho)_(delta)^(sigma)+R_(alpha rho delta sigma)R_(beta)^(rho)gamma^(sigma)-(1)/(8)g_(alpha beta)g_(gamma delta)R_(rho sigma lambda mu)R^(rho sigma lambda mu).:}\begin{equation*} T_{\alpha \beta \gamma \delta}=R_{\alpha \rho \gamma \sigma} R_{\beta}{ }^{\rho}{ }_{\delta}{ }^{\sigma}+R_{\alpha \rho \delta \sigma} R_{\beta}{ }^{\rho} \gamma^{\sigma}-\frac{1}{8} g_{\alpha \beta} g_{\gamma \delta} R_{\rho \sigma \lambda \mu} R^{\rho \sigma \lambda \mu} . \tag{15.30a} \end{equation*}(15.30a)Tαβγδ=RαργσRβρδσ+RαρδσRβργσ18gαβgγδRρσλμRρσλμ.
Show also that in empty spacetime
(15.30b) T α β γ δ ; α = 0 , (15.30b) T α β γ δ ; α = 0 , {:(15.30b)T^(alpha)_(beta gamma delta;alpha)=0",":}\begin{equation*} T^{\alpha}{ }_{\beta \gamma \delta ; \alpha}=0, \tag{15.30b} \end{equation*}(15.30b)Tαβγδ;α=0,
T α β γ δ T α β γ δ T_(alpha beta gamma delta)T_{\alpha \beta \gamma \delta}Tαβγδ is symmetric and traceless on all pairs of indices.
(15.30c) T α β γ δ is symmetric and traceless on all pairs of indices. (15.30c) T α β γ δ  is symmetric and traceless on all pairs of indices.  {:(15.30c)T_(alpha beta gamma delta)" is symmetric and traceless on all pairs of indices. ":}\begin{equation*} T_{\alpha \beta \gamma \delta} \text { is symmetric and traceless on all pairs of indices. } \tag{15.30c} \end{equation*}(15.30c)Tαβγδ is symmetric and traceless on all pairs of indices. 
Discussion: It turns out that Einstein's "canonical energy-momentum pseudotensor" ( $ 20.3 $ 20.3 $20.3\$ 20.3$20.3 ) for the gravitational field in empty spacetime has a second derivative which, in a Riemannnormal coordinate system, is
(15.31a) t E α β , γ δ = 4 9 ( T α β γ δ 1 4 S α β γ δ ) . (15.31a) t E α β , γ δ = 4 9 T α β γ δ 1 4 S α β γ δ . {:(15.31a)t_(Ealpha beta,gamma delta)=-(4)/(9)(T_(alpha beta gamma delta)-(1)/(4)S_(alpha beta gamma delta)).:}\begin{equation*} t_{\mathrm{E} \alpha \beta, \gamma \delta}=-\frac{4}{9}\left(T_{\alpha \beta \gamma \delta}-\frac{1}{4} S_{\alpha \beta \gamma \delta}\right) . \tag{15.31a} \end{equation*}(15.31a)tEαβ,γδ=49(Tαβγδ14Sαβγδ).
Here T α β γ δ T α β γ δ T_(alpha beta gamma delta)T_{\alpha \beta \gamma \delta}Tαβγδ is the completely symmetric Bel-Robinson tensor, and S α β γ δ S α β γ δ S_(alpha beta gamma delta)S_{\alpha \beta \gamma \delta}Sαβγδ is defined by
(15.31b) S α β γ δ R α δ ρ σ R β γ ρ σ + R α γ ρ σ R β δ ρ σ + 1 4 g α β g γ δ R μ ν ρ σ R μ ν ρ σ . (15.31b) S α β γ δ R α δ ρ σ R β γ ρ σ + R α γ ρ σ R β δ ρ σ + 1 4 g α β g γ δ R μ ν ρ σ R μ ν ρ σ . {:(15.31b)S_(alpha beta gamma delta)-=R_(alpha delta rho sigma)R_(beta gamma)^(rho sigma)+R_(alpha gamma rho sigma)R_(beta delta)^(rho sigma)+(1)/(4)g_(alpha beta)g_(gamma delta)R_(mu nu rho sigma)R^(mu nu rho sigma).:}\begin{equation*} S_{\alpha \beta \gamma \delta} \equiv R_{\alpha \delta \rho \sigma} R_{\beta \gamma}{ }^{\rho \sigma}+R_{\alpha \gamma \rho \sigma} R_{\beta \delta}{ }^{\rho \sigma}+\frac{1}{4} g_{\alpha \beta} g_{\gamma \delta} R_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma} . \tag{15.31b} \end{equation*}(15.31b)SαβγδRαδρσRβγρσ+RαγρσRβδρσ+14gαβgγδRμνρσRμνρσ.
S α β γ δ S α β γ δ S_(alpha beta gamma delta)S_{\alpha \beta \gamma \delta}Sαβγδ appears in the empty-space covariant wave equation
(15.31c) Δ R α β γ δ R α β γ δ ; μ μ + R α β ρ σ R γ δ ρ σ + 2 ( R α ρ γ σ R β ρ δ σ R α ρ δ σ R β ρ σ σ ) = 0 , (15.31c) Δ R α β γ δ R α β γ δ ; μ μ + R α β ρ σ R γ δ ρ σ + 2 R α ρ γ σ R β ρ δ σ R α ρ δ σ R β ρ σ σ = 0 , {:(15.31c)DeltaR_(alpha beta gamma delta)-=-R_(alpha beta gamma delta;mu^('mu))+R_(alpha beta rho sigma)R_(gamma delta)^(rho sigma)+2(R_(alpha rho gamma sigma)R_(beta)^(rho)_(delta)^(sigma)-R_(alpha rho delta sigma)R_(beta)^(rho)^(sigma)^(sigma))=0",":}\begin{equation*} \Delta R_{\alpha \beta \gamma \delta} \equiv-R_{\alpha \beta \gamma \delta ; \mu^{\prime \mu}}+R_{\alpha \beta \rho \sigma} R_{\gamma \delta}{ }^{\rho \sigma}+2\left(R_{\alpha \rho \gamma \sigma} R_{\beta}{ }^{\rho}{ }_{\delta}{ }^{\sigma}-R_{\alpha \rho \delta \sigma} R_{\beta}{ }^{\rho}{ }^{\sigma}{ }^{\sigma}\right)=0, \tag{15.31c} \end{equation*}(15.31c)ΔRαβγδRαβγδ;μμ+RαβρσRγδρσ+2(RαργσRβρδσRαρδσRβρσσ)=0,
where Δ Δ Delta\DeltaΔ is a variant of the Lichnerowicz-de Rham wave operator [Lichnerowicz (1964)], when one rewrites this wave equation as
(15.31d) R α β γ δ R α β γ δ ; μ ; μ = 2 S [ α [ γ β ] δ ] . (15.31d) R α β γ δ R α β γ δ ; μ ; μ = 2 S [ α [ γ β ] δ ] . {:(15.31d)◻R_(alpha beta)^(gamma delta)-=R_(alpha beta)^(gamma delta)_(;mu)^(;mu)=2S_([alpha)^([gamma)_(beta])^(delta]).:}\begin{equation*} \square R_{\alpha \beta}{ }^{\gamma \delta} \equiv R_{\alpha \beta}{ }^{\gamma \delta}{ }_{; \mu}{ }^{; \mu}=2 S_{[\alpha}{ }^{[\gamma}{ }_{\beta]}{ }^{\delta]} . \tag{15.31d} \end{equation*}(15.31d)RαβγδRαβγδ;μ;μ=2S[α[γβ]δ].

EINSTEIN'S GEOMETRIC THEORY OF GRAVITY

Wherein the reader is seduced into marriage with the most elegant temptress of all-Geometrodynamics - and learns from her the magic potions and incantations that control the universe.

cumera 16

EQUIVALENCE PRINCIPLE AND MEASUREMENT OF THE "'GRAVITATIONAL FIELD"

Rather than have one global frame with gravitational forces we have many local frames without gravitational forces.

STEPHEN SCHUTZ (1966)

§16.1. OVERVIEW

With the mathematics of curved spacetime now firmly in hand, one is tempted to rush headlong into a detailed study of Einstein's field equations. But such temptation must be resisted for a short time more. To grasp the field equations fully, one must first understand how the classical laws of physics change, or do not change, in the transition from flat spacetime to curved ( § § 16.2 § § 16.2 §§16.2\S \S 16.2§§16.2 and 16.3 ); and one must understand how the "gravitational field" (metric; covariant derivative; spacetime curvature; . . .) can be "measured" ( § 16.4 § 16.4 §16.4\S 16.4§16.4 and 16.5).

§16.2. THE LAWS OF PHYSICS IN CURVED SPACETIME

Wherever one is and whenever one probes, one finds that then and there one can introduce a local inertial frame in which all test particles move along straight lines. Moreover, this local inertial frame is also locally Lorentz: in it the velocity of light has its standard value, and light rays, like world lines of test particles, are straight. But physics is more, and the analysis of physics demands more than an account solely of the motions of test particles and light rays. What happens to Maxwell's equations, the laws of hydrodynamics, the principles of atomic structure, and all the rest of physics under the influence of "powerful gravitational fields"?
Purpose of this chapter
Einstein's equivalence principle
Equivalence principle as tool to mesh nongravitational laws with gravity
The answer is simple: in any and every local Lorentz frame, anywhere and anytime in the universe, all the (nongravitational) laws of physics must take on their familiar special-relativistic forms. Equivalently: there is no way, by experiments confined to infinitesimally small regions of spacetime, to distinguish one local Lorentz frame in one region of spacetime from any other local Lorentz frame in the same or any other region. This is Einstein's principle of equivalence in its strongest form-a principle that is compelling both philosophically and experimentally. (For the relevant experimental tests, see §38.6.)
The principle of equivalence has great power. With it one can generalize all the special relativistic laws of physics to curved spacetime. And the curvature need not be small. It may be as large as that in the center of a neutron star; as large as that at the edge of a black hole; arbitrarily large, in fact-or almost so. Only at the endpoint of gravitational collapse and in the initial instant of the "big bang," i.e., only at "singularities of spacetime," will there be a breakdown in the conditions needed for direct application of the equivalence principle (see § § 28.3 , 34.6 , 43.3 § § 28.3 , 34.6 , 43.3 §§28.3,34.6,43.3\S \S 28.3,34.6,43.3§§28.3,34.6,43.3, 43.4, and chapter 44). Everywhere else the equivalence principle acts as a tool to mesh all the nongravitational laws of physics with gravity.
Example: Mesh the "law of local energy-momentum conservation," T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0, with gravity. Solution:
(1) The law in flat spacetime, written in abstract geometric form, reads
(16.1a) T = 0 (16.1a) T = 0 {:(16.1a)grad*T=0:}\begin{equation*} \boldsymbol{\nabla} \cdot \boldsymbol{T}=0 \tag{16.1a} \end{equation*}(16.1a)T=0
(2) Rewritten in a global Lorentz frame of flat spacetime, it reads
(16.1b) T μ v , ν = 0 . (16.1b) T μ v , ν = 0 . {:(16.1b)T^(mu v)_(,nu)=0.:}\begin{equation*} T^{\mu v}{ }_{, \nu}=0 . \tag{16.1b} \end{equation*}(16.1b)Tμv,ν=0.
(3) Application of equivalence principle gives same equation in local Lorentz frame of curved spacetime:
(16.1c) T μ ^ ν ^ ν ^ , ν ^ = 0 at origin of local Lorentz frame. (16.1c) T μ ^ ν ^ ν ^ , ν ^ = 0  at origin of local Lorentz frame.  {:(16.1c)T^( hat(mu) hat(nu) hat(nu), hat(nu))=0" at origin of local Lorentz frame. ":}\begin{equation*} T^{\hat{\mu} \hat{\nu} \hat{\nu}, \hat{\nu}}=0 \text { at origin of local Lorentz frame. } \tag{16.1c} \end{equation*}(16.1c)Tμ^ν^ν^,ν^=0 at origin of local Lorentz frame. 
Because the connection coefficients vanish at the origin of the local Lorentz frame, this can be rewritten as
(16.1d) T μ ^ ν ^ ; v ^ = 0 at origin of local Lorentz frame. (16.1d) T μ ^ ν ^ ; v ^ = 0  at origin of local Lorentz frame.  {:(16.1d)T^( hat(mu) hat(nu))_(; hat(v))=0" at origin of local Lorentz frame. ":}\begin{equation*} T^{\hat{\mu} \hat{\nu}}{ }_{; \hat{v}}=0 \text { at origin of local Lorentz frame. } \tag{16.1d} \end{equation*}(16.1d)Tμ^ν^;v^=0 at origin of local Lorentz frame. 
(4) The geometric law in curved spacetime, of which these are the local-Lorentz components, is
(16.1e) T = 0 (16.1e) T = 0 {:(16.1e)grad*T=0:}\begin{equation*} \boldsymbol{\nabla} \cdot \boldsymbol{T}=0 \tag{16.1e} \end{equation*}(16.1e)T=0
and its component formulation in any reference frame reads
(16.1f) T μ ν ; ν = 0 . (16.1f) T μ ν ; ν = 0 . {:(16.1f)T^(mu nu)_(;nu)=0.:}\begin{equation*} T^{\mu \nu}{ }_{; \nu}=0 . \tag{16.1f} \end{equation*}(16.1f)Tμν;ν=0.
Compare the abstract geometric law (16.1e) in curved spacetime with the corresponding law (16.1a) in flat spacetime. They are identical! That this is not an accident one can readily see by tracing out the above four-step argument for any other law
of physics (e.g., Maxwell's equation F = 4 π J F = 4 π J grad*F=4pi J\boldsymbol{\nabla} \cdot \boldsymbol{F}=4 \pi \boldsymbol{J}F=4πJ ). The laws of physics, written in abstract geometric form, differ in no way whatsoever between curved spacetime and flat spacetime; this is guaranteed by, and in fact is a mere rewording of, the equivalence principle.
Compare the component version of the law T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0, as written in an arbitrary frame in curved spacetime [equation (16.1f)], with the component version in a global Lorentz frame of flat spacetime [equation (16.1b)]. They differ in only one way: the comma (partial derivative; flat-spacetime gradient) is replaced by a semicolon (covariant derivative; curved-spacetime gradient). This procedure for rewriting the equations has universal application. The laws of physics, written in component form, change on passage from flat spacetime to curved spacetime by a mere replacement of all commas by semicolons (no change at all physically or geometrically; change due only to switch in reference frame from Lorentz to non-Lorentz!). This statement, like the nonchanging of abstract geometric laws, is nothing but a rephrased version of the equivalence principle.
The transition in formalism from flat spacetime to curved spacetime is a trivial process when performed as outlined above. But it is nontrivial in its implications. It meshes gravity with all the laws of physics. Gravity enters in an essential way through the covariant derivative of curved spacetime, as one sees clearly in the following exercise.

"Comma-goes-to-semicolon"

rule

Exercise 16.1. HYDRODYNAMICS IN A WEAK GRAVITATIONAL FIELD

(a) In § 18.4 § 18.4 §18.4\S 18.4§18.4 it will be shown that for a nearly Newtonian system, analyzed in an appropriate nearly global Lorentz coordinate system, the metric has the form
(16.2a) d s 2 = ( 1 + 2 Φ ) d t 2 + ( 1 2 Φ ) ( d x 2 + d y 2 + d z 2 ) (16.2a) d s 2 = ( 1 + 2 Φ ) d t 2 + ( 1 2 Φ ) d x 2 + d y 2 + d z 2 {:(16.2a)ds^(2)=-(1+2Phi)dt^(2)+(1-2Phi)(dx^(2)+dy^(2)+dz^(2)):}\begin{equation*} d s^{2}=-(1+2 \Phi) d t^{2}+(1-2 \Phi)\left(d x^{2}+d y^{2}+d z^{2}\right) \tag{16.2a} \end{equation*}(16.2a)ds2=(1+2Φ)dt2+(12Φ)(dx2+dy2+dz2)
where Φ Φ Phi\PhiΦ is the Newtonian potential ( 1 Φ < 0 ) ( 1 Φ < 0 ) (-1≪Phi < 0)(-1 \ll \Phi<0)(1Φ<0). Consider a nearly Newtonian perfect fluid [stress-energy tensor
(16.2b) T α β = ( ρ + p ) u α u β + p g α β , p ρ ; (16.2b) T α β = ( ρ + p ) u α u β + p g α β , p ρ ; {:(16.2b)T^(alpha beta)=(rho+p)u^(alpha)u^(beta)+pg^(alpha beta)","quad p≪rho;:}\begin{equation*} T^{\alpha \beta}=(\rho+p) u^{\alpha} u^{\beta}+p g^{\alpha \beta}, \quad p \ll \rho ; \tag{16.2b} \end{equation*}(16.2b)Tαβ=(ρ+p)uαuβ+pgαβ,pρ;
see Box 5.1 and § 5.10 § 5.10 §5.10\S 5.10§5.10 ] moving in such a spacetime with ordinary velocity
(16.2c) v j d x j / d t 1 (16.2c) v j d x j / d t 1 {:(16.2c)v^(j)-=dx^(j)//dt≪1:}\begin{equation*} v^{j} \equiv d x^{j} / d t \ll 1 \tag{16.2c} \end{equation*}(16.2c)vjdxj/dt1
Show that the equations T μ ν ; ν = 0 T μ ν ; ν = 0 T^(mu nu)_(;nu)=0T^{\mu \nu}{ }_{; \nu}=0Tμν;ν=0 for this system reduce to the familiar Newtonian law of mass conservation, and the Newtonian equation of motion for a fluid in a gravitational field:
(16.3a) d ρ d t = ρ v j x j , ρ d v j d t = ρ Φ x j p x j (16.3a) d ρ d t = ρ v j x j , ρ d v j d t = ρ Φ x j p x j {:(16.3a)(d rho)/(dt)=-rho(delv^(j))/(delx^(j))","quad rho(dv^(j))/(dt)=-rho(del Phi)/(delx^(j))-(del p)/(delx^(j)):}\begin{equation*} \frac{d \rho}{d t}=-\rho \frac{\partial v^{j}}{\partial x^{j}}, \quad \rho \frac{d v^{j}}{d t}=-\rho \frac{\partial \Phi}{\partial x^{j}}-\frac{\partial p}{\partial x^{j}} \tag{16.3a} \end{equation*}(16.3a)dρdt=ρvjxj,ρdvjdt=ρΦxjpxj
where d / d t d / d t d//dtd / d td/dt is the time derivative comoving with the matter
(16.3b) d d t t + v j x j (16.3b) d d t t + v j x j {:(16.3b)(d)/(dt)-=(del)/(del t)+v^(j)(del)/(delx^(j)):}\begin{equation*} \frac{d}{d t} \equiv \frac{\partial}{\partial t}+v^{j} \frac{\partial}{\partial x^{j}} \tag{16.3b} \end{equation*}(16.3b)ddtt+vjxj

EXERCISES

(b) Use these equations to calculate the pressure gradient in the Earth's atmosphere as a function of temperature and pressure. In the calculation, use the nonrelativistic relation ρ = n M μ M ρ = n M μ M rho=n_(M)mu_(M)\rho=n_{M} \mu_{M}ρ=nMμM, where n M n M n_(M)n_{M}nM is the number density of molecules and μ M μ M mu_(M)\mu_{M}μM is the mean rest mass per molecule; use the ideal-gas equation of state
p = n M k T ( k = Boltzmann's constant ) p = n M k T ( k =  Boltzmann's constant  ) p=n_(M)kT quad(k=" Boltzmann's constant ")p=n_{M} k T \quad(k=\text { Boltzmann's constant })p=nMkT(k= Boltzmann's constant )
and use the spherically symmetric form, Φ = M / r Φ = M / r Phi=-M//r\Phi=-M / rΦ=M/r, for the Earth's Newtonian potential. If the pressure at sea level is 1.01 × 10 6 1.01 × 10 6 1.01 xx10^(6)1.01 \times 10^{6}1.01×106 dynes / cm 2 / cm 2 //cm^(2)/ \mathrm{cm}^{2}/cm2, what, approximately, is the pressure on top of Mount Everest (altitude 8,840 meters)? (Make a reasonable assumption about the temperature distribution of the atmosphere.)

Exercise 16.2. WORLD LINES OF PHOTONS

Show that in flat spacetime the conservation law for the 4 -momentum of a freely moving photon can be written
(16.4a) p p = 0 (16.4a) p p = 0 {:(16.4a)grad_(p)p=0:}\begin{equation*} \nabla_{p} p=0 \tag{16.4a} \end{equation*}(16.4a)pp=0
According to the equivalence principle, this equation must be true also in curved spacetime. Show that this means photons move along null geodesics of curved spacetime with affine parameter λ λ lambda\lambdaλ related to 4 -momentum by
(16.4b) p = d / d λ (16.4b) p = d / d λ {:(16.4b)p=d//d lambda:}\begin{equation*} \boldsymbol{p}=d / d \lambda \tag{16.4b} \end{equation*}(16.4b)p=d/dλ
In exercise 18.6 this result will be used to calculate the deflection of light by the sun.
Factor-ordering problems and coupling to curvature

§16.3. FACTOR-ORDERING PROBLEMS IN THE EQUIVALENCE PRINCIPLE

On occasion in applying the equivalence principle to get from physics in flat spacetime to physics in curved spacetime one encounters "factor-ordering problems" analogous to those that beset the transition from classical mechanics to quantum mechanics.* Example: How is the equation (3.56) for the vector potential of electrodynamics to be translated into curved spacetime? If the flat-spacetime equation is written
A α , μ μ + A , μ μ α = 4 π J α , A α , μ μ + A , μ μ α = 4 π J α , -A^(alpha,mu)_(mu)+A_(,mu)^(mu)^(alpha)=4piJ^(alpha),-A^{\alpha, \mu}{ }_{\mu}+A_{, \mu}^{\mu}{ }^{\alpha}=4 \pi J^{\alpha},Aα,μμ+A,μμα=4πJα,
then its transition ("comma goes to semicolon") reads
(16.5) A α ; μ + A ; μ μ = 4 π J α (16.5) A α ; μ + A ; μ μ = 4 π J α {:(16.5)-A^(alpha;mu)+A_(;mu)^(mu)=4piJ^(alpha):}\begin{equation*} -A^{\alpha ; \mu}+A_{; \mu}^{\mu}=4 \pi J^{\alpha} \tag{16.5} \end{equation*}(16.5)Aα;μ+A;μμ=4πJα
However, if the flat-spacetime equation is written with two of its partial derivatives interchanged
A α , μ μ + A μ , α μ = 4 π J α A α , μ μ + A μ , α μ = 4 π J α -A^(alpha,mu)_(mu)+A^(mu,alpha)_(mu)=4piJ^(alpha)-A^{\alpha, \mu}{ }_{\mu}+A^{\mu, \alpha}{ }_{\mu}=4 \pi J^{\alpha}Aα,μμ+Aμ,αμ=4πJα
then its translation reads
A α ; μ μ + A μ ; α μ = 4 π J α , A α ; μ μ + A μ ; α μ = 4 π J α , -A^(alpha;mu)_(mu)+A^(mu;alpha)_(mu)=4piJ^(alpha),-A^{\alpha ; \mu}{ }_{\mu}+A^{\mu ; \alpha}{ }_{\mu}=4 \pi J^{\alpha},Aα;μμ+Aμ;αμ=4πJα,
which can be rewritten
$$
( ) A α ; μ μ + A μ ; μ α + R α μ A μ = 4 π J α . ( ) A α ; μ μ + A μ ; μ α + R α μ A μ = 4 π J α . {:('")"-A^(alpha;mu)_(mu)+A^(mu)_(;mu)^(alpha)+R^(alpha)_(mu)A^(mu)=4piJ^(alpha).:}\begin{equation*} -A^{\alpha ; \mu}{ }_{\mu}+A^{\mu}{ }_{; \mu}{ }^{\alpha}+R^{\alpha}{ }_{\mu} A^{\mu}=4 \pi J^{\alpha} . \tag{$\prime$} \end{equation*}()Aα;μμ+Aμ;μα+RαμAμ=4πJα.
$$
(Ricci tensor appears as result of interchanging covariant derivatives; see exercise 16.3.) Which equation is correct-(16.5) or ( 16.5 16.5 16.5^(')16.5^{\prime}16.5 )? This question is nontrivial, just as the analogous factor-ordering problems of quantum theory are nontrivial. For rules-of-thumb that resolve this and most factor-ordering problems, see Box 16.1. These rules tell one that ( 16.5 16.5 16.5^(')16.5^{\prime}16.5 ) is correct and (16.5) is wrong (see Box 16.1 and §22.4).

Exercise 16.3. NONCOMMUTATION OF COVARIANT DERIVATIVES

Let B B B\boldsymbol{B}B be a vector field and S S S\boldsymbol{S}S be a second-rank tensor field. Show that
(16.6a) B ; α β μ = B ; β α μ + R ν β α μ B ν (16.6b) S ; α β μ ν = S ; β α μ ν + R μ ρ β α S ρ ν + R ρ β α v S μ ρ . (16.6a) B ; α β μ = B ; β α μ + R ν β α μ B ν (16.6b) S ; α β μ ν = S ; β α μ ν + R μ ρ β α S ρ ν + R ρ β α v S μ ρ . {:[(16.6a)B_(;alpha beta)^(mu)=B_(;beta alpha)^(mu)+R_(nu beta alpha)^(mu)B^(nu)],[(16.6b)S_(;alpha beta)^(mu nu)=S_(;beta alpha)^(mu nu)+R^(mu)_(rho beta alpha)S^(rho nu)+R_(rho beta alpha)^(v)S^(mu rho).]:}\begin{align*} B_{; \alpha \beta}^{\mu} & =B_{; \beta \alpha}^{\mu}+R_{\nu \beta \alpha}^{\mu} B^{\nu} \tag{16.6a}\\ S_{; \alpha \beta}^{\mu \nu} & =S_{; \beta \alpha}^{\mu \nu}+R^{\mu}{ }_{\rho \beta \alpha} S^{\rho \nu}+R_{\rho \beta \alpha}^{v} S^{\mu \rho} . \tag{16.6b} \end{align*}(16.6a)B;αβμ=B;βαμ+RνβαμBν(16.6b)S;αβμν=S;βαμν+RμρβαSρν+RρβαvSμρ.
From equation (16.6a), show that
(16.6c) B μ ; α μ = B μ ; μ α + R α μ B μ . (16.6c) B μ ; α μ = B μ ; μ α + R α μ B μ . {:(16.6c)B^(mu;alpha)_(mu)=B^(mu)_(;mu)^(alpha)+R^(alpha)_(mu)B^(mu).:}\begin{equation*} B^{\mu ; \alpha}{ }_{\mu}=B^{\mu}{ }_{; \mu}{ }^{\alpha}+R^{\alpha}{ }_{\mu} B^{\mu} . \tag{16.6c} \end{equation*}(16.6c)Bμ;αμ=Bμ;μα+RαμBμ.
[Hint for Track-1 calculation: Work in a local Lorentz frame, where Γ α β γ = 0 Γ α β γ = 0 Gamma^(alpha)_(beta gamma)=0\Gamma^{\alpha}{ }_{\beta \gamma}=0Γαβγ=0 but Γ α β γ , δ 0 Γ α β γ , δ 0 Gamma^(alpha)_(beta gamma,delta)!=0\Gamma^{\alpha}{ }_{\beta \gamma, \delta} \neq 0Γαβγ,δ0; expand the lefthand side in terms of Christoffel symbols and partial derivatives; and use equation (8.44) for the Riemann tensor. An alternative Track-2 calculation notices that β α B β α B grad_(beta)grad_(alpha)B\boldsymbol{\nabla}_{\beta} \boldsymbol{\nabla}_{\alpha} \boldsymbol{B}βαB is not linear in e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα, and that B μ ; α β B μ ; α β B^(mu)_(;alpha beta)B^{\mu}{ }_{; \alpha \beta}Bμ;αβ are not its components; but, rather, that
(16.7) B ; α β μ 4 B ( ω μ , e α , e β ) . (16.7) B ; α β μ 4 B ω μ , e α , e β . {:(16.7)B_(;alpha beta)^(mu)-=_(_(4)^(grad)grad B(omega^(mu),e_(alpha),e_(beta)).):}\begin{equation*} B_{; \alpha \beta}^{\mu} \equiv \underset{{ }_{4}^{\boldsymbol{\nabla}} \boldsymbol{\nabla} \boldsymbol{B}\left(\boldsymbol{\omega}^{\mu}, \boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right) .}{ } \tag{16.7} \end{equation*}(16.7)B;αβμ4B(ωμ,eα,eβ).
The calculation then proceeds as follows:
ω μ , β α B = ω μ , β ( e α B ) = ω μ , ( β e α ) B + e α ( β B ) = ω μ , Γ v α β e v B + B ( , e α , e β ) = B μ ; ν Γ v α β + B μ ; α β . ω μ , β α B = ω μ , β e α B = ω μ , β e α B + e α β B = ω μ , Γ v α β e v B + B , e α , e β = B μ ; ν Γ v α β + B μ ; α β . {:[(:omega^(mu),grad_(beta)grad_(alpha)B:)=(:omega^(mu),grad_(beta)(e_(alpha)*grad B):)],[=(:omega^(mu),(grad_(beta)e_(alpha))*grad B+e_(alpha)*(grad_(beta)grad B):)],[=(:omega^(mu),Gamma^(v)_(alpha beta)e_(v)*grad B+grad grad B(dots,e_(alpha),e_(beta)):)],[=B^(mu)_(;nu)Gamma^(v)_(alpha beta)+B^(mu)_(;alpha beta).]:}\begin{aligned} \left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{\nabla}_{\beta} \boldsymbol{\nabla}_{\alpha} \boldsymbol{B}\right\rangle & =\left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{\nabla}_{\beta}\left(\boldsymbol{e}_{\alpha} \cdot \boldsymbol{\nabla} \boldsymbol{B}\right)\right\rangle \\ & =\left\langle\boldsymbol{\omega}^{\mu},\left(\boldsymbol{\nabla}_{\beta} \boldsymbol{e}_{\alpha}\right) \cdot \boldsymbol{\nabla} \boldsymbol{B}+\boldsymbol{e}_{\alpha} \cdot\left(\boldsymbol{\nabla}_{\beta} \boldsymbol{\nabla} \boldsymbol{B}\right)\right\rangle \\ & =\left\langle\boldsymbol{\omega}^{\mu}, \Gamma^{v}{ }_{\alpha \beta} \boldsymbol{e}_{v} \cdot \boldsymbol{\nabla} \boldsymbol{B}+\boldsymbol{\nabla} \boldsymbol{\nabla} \boldsymbol{B}\left(\ldots, \boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right)\right\rangle \\ & ={B^{\mu}}_{; \nu} \Gamma^{v}{ }_{\alpha \beta}+B^{\mu}{ }_{; \alpha \beta} . \end{aligned}ωμ,βαB=ωμ,β(eαB)=ωμ,(βeα)B+eα(βB)=ωμ,ΓvαβevB+B(,eα,eβ)=Bμ;νΓvαβ+Bμ;αβ.
Consequently
B μ ; α β B μ ; β α = ω μ , [ β , α ] B B μ ; ν ( Γ v α β Γ ν β α ) = ω μ , [ β , α ] B ω μ , ( β e α e e β ) B = = ω μ , ( [ β , α ] [ e β , e α ) ) B ω μ , R ( e β , e α ) B = R μ ν β α B v , B μ ; α β B μ ; β α = ω μ , β , α B B μ ; ν Γ v α β Γ ν β α = ω μ , β , α B ω μ , β e α e e β B = = ω μ , β , α e β , e α B ω μ , R e β , e α B = R μ ν β α B v , {:[B^(mu)_(;alpha beta)-B^(mu)_(;beta alpha)=(:omega^(mu),[grad_(beta),grad_(alpha)]B:)-B^(mu)_(;nu)(Gamma^(v)_(alpha beta)-Gamma^(nu)_(beta alpha))],[=(:omega^(mu),[grad_(beta),grad_(alpha)]B:)-(:omega^(mu),grad_({:(grad_(beta)e_(alpha)-grad_(e)e_(beta))B:))^(=):}],[{:=(:omega^(mu),([grad_(beta),grad_(alpha)]-grad_([e_(beta),e_(alpha))))B:)(^(omega^(mu))),R(e_(beta),e_(alpha))B:)],[=R^(mu)_(nu beta alpha)B^(v)","]:}\begin{aligned} B^{\mu}{ }_{; \alpha \beta}-B^{\mu}{ }_{; \beta \alpha} & =\left\langle\boldsymbol{\omega}^{\mu},\left[\boldsymbol{\nabla}_{\beta}, \boldsymbol{\nabla}_{\alpha}\right] \boldsymbol{B}\right\rangle-B^{\mu}{ }_{; \nu}\left(\Gamma^{v}{ }_{\alpha \beta}-\Gamma^{\nu}{ }_{\beta \alpha}\right) \\ & =\left\langle\boldsymbol{\omega}^{\mu},\left[\boldsymbol{\nabla}_{\beta}, \boldsymbol{\nabla}_{\alpha}\right] \boldsymbol{B}\right\rangle-\left\langle\boldsymbol{\omega}^{\mu}, \mathbf{\nabla}_{\left.\left(\boldsymbol{\nabla}_{\beta} \boldsymbol{e}_{\alpha}-\mathbf{\nabla}_{e} \boldsymbol{e}_{\beta}\right) \boldsymbol{B}\right\rangle}^{=}\right. \\ & \left.=\left\langle\boldsymbol{\omega}^{\mu},\left(\left[\boldsymbol{\nabla}_{\beta}, \boldsymbol{\nabla}_{\alpha}\right]-\boldsymbol{\nabla}_{\left[\boldsymbol{e}_{\beta}, \boldsymbol{e}_{\alpha}\right)}\right) \boldsymbol{B}\right\rangle \stackrel{\boldsymbol{\omega}^{\mu}}{ }, \mathscr{R}\left(\boldsymbol{e}_{\beta}, \boldsymbol{e}_{\alpha}\right) \boldsymbol{B}\right\rangle \\ & =R^{\mu}{ }_{\nu \beta \alpha} \boldsymbol{B}^{v}, \end{aligned}Bμ;αβBμ;βα=ωμ,[β,α]BBμ;ν(ΓvαβΓνβα)=ωμ,[β,α]Bωμ,(βeαeeβ)B==ωμ,([β,α][eβ,eα))Bωμ,R(eβ,eα)B=RμνβαBv,
in agreement with (16.6a). Note: because of slight ambiguity in the abstract notation, one must think carefully about each step in the above calculation. Component notation, by contrast, is completely unambiguous.]

Box 16.1 FACTOR ORDERING AND CURVATURE COUPLING IN APPLICATIONS OF THE EQUIVALENCE PRINCIPLE

The Problem

In what order should derivatives be written when applying the "comma-goes-tosemicolon rule"? Interchanging derivatives makes no difference in flat spacetime, but in curved spacetime it produces terms that couple to curvature, e.g., 2 B α ; [ γ β ] 2 B α ; [ γ β ] 2B^(alpha)_(;[gamma beta])-=2 B^{\alpha}{ }_{;[\gamma \beta]} \equiv2Bα;[γβ] B α ; γ β B α ; β γ = R α μ β γ B μ B α ; γ β B α ; β γ = R α μ β γ B μ B^(alpha)_(;gamma beta)-B^(alpha)_(;beta gamma)=R^(alpha)_(mu beta gamma)B^(mu)B^{\alpha}{ }_{; \gamma \beta}-B^{\alpha}{ }_{; \beta \gamma}=R^{\alpha}{ }_{\mu \beta \gamma} B^{\mu}Bα;γβBα;βγ=RαμβγBμ for any vector field (see exercise 16.3). Hence, the problem can be restated: When must the comma-goes-to-semicolon rule be augmented by terms that couple to curvature?

The Solution

There is no solution in general, but in most cases the following types of mathematical and physical reasoning resolve the problem unambiguously.
A. Mathematically, curvature terms almost always arise from the noncommutation of covariant derivatives. Consequently, one needs to worry about curvature terms in any equation that contains a double covariant derivative (e.g., A α , μ μ + A μ , μ α = A α , μ μ + A μ , μ α = -A^(alpha,mu)_(mu)+A^(mu)_(,mu)^(alpha)=-A^{\alpha, \mu}{ }_{\mu}+A^{\mu}{ }_{, \mu}{ }^{\alpha}=Aα,μμ+Aμ,μα= 4 π J α 4 π J α 4piJ^(alpha)4 \pi J^{\alpha}4πJα ); or in any equation whose derivation from more fundamental laws involves double covariant derivatives (e.g. μ S = 0 μ S = 0 grad_(mu)S=0\boldsymbol{\nabla}_{\boldsymbol{\mu}} \boldsymbol{S}=0μS=0 in Example B(3) below). But one can ignore curvature coupling everywhere else (e.g., in Maxwell's first-order equations).
B. Coupling to curvature can surely not occur without some physical reason. Therefore, if one applies the comma-goes-to-semicolon rule only to physically measurable quantities (e.g., to the electromagnetic field, but not to the vector potential), one can "intuit" whether coupling to curvature is likely. Examples:
(1) Local energy-momentum conservation. A coupling to curvature in the equations T α β ; β = 0 T α β ; β = 0 T^(alpha beta)_(;beta)=0T^{\alpha \beta}{ }_{; \beta}=0Tαβ;β=0-e.g., replacing them by T α β ; β = R α β γ δ T β γ u δ T α β ; β = R α β γ δ T β γ u δ T^(alpha beta)_(;beta)=R^(alpha)_(beta gamma delta)T^(beta gamma)u^(delta)T^{\alpha \beta}{ }_{; \beta}=R^{\alpha}{ }_{\beta \gamma \delta} T^{\beta \gamma} u^{\delta}Tαβ;β=RαβγδTβγuδ-would not make sense at all. In a local inertial frame such terms as R α β γ δ T β γ u δ R α β γ δ T β γ u δ R^(alpha)_(beta gamma delta)T^(beta gamma)u^(delta)R^{\alpha}{ }_{\beta \gamma \delta} T^{\beta \gamma} u^{\delta}RαβγδTβγuδ would be interpreted as forces produced at a single point by curvature. But it should not be possible to feel curvature except over finite regions (geodesic deviation, etc.)! Put differently, the second derivatives of the gravitational potential (metric) can hardly produce net forces at a point; they should only produce tidal forces!
(2) Maxwell's equations for the electromagnetic field tensor. Here it would also be unnatural to introduce curvature terms. They would cause a breakdown in charge conservation, in the sense of termination of electric and magnetic field lines at points where there is curvature but no charge. To maintain charge conservation, one omits curvature coupling when one translates Maxwell's equations (3.32) and (3.36) into curved spacetime:
F ; β α β = 4 π J α , F α β ; γ + F β γ ; α + F γ α ; β = 0 . F ; β α β = 4 π J α , F α β ; γ + F β γ ; α + F γ α ; β = 0 . F_(;beta)^(alpha beta)=4piJ^(alpha),quadF_(alpha beta;gamma)+F_(beta gamma;alpha)+F_(gamma alpha;beta)=0.F_{; \beta}^{\alpha \beta}=4 \pi J^{\alpha}, \quad F_{\alpha \beta ; \gamma}+F_{\beta \gamma ; \alpha}+F_{\gamma \alpha ; \beta}=0 .F;βαβ=4πJα,Fαβ;γ+Fβγ;α+Fγα;β=0.
Moreover, one continues to regard F μ ν F μ ν F_(mu nu)F_{\mu \nu}Fμν as arising from a vector potential by the curved-spacetime translation of ( 3.54 3.54 3.54^(')3.54^{\prime}3.54 )
F μ ν = A ν ; μ A μ ; ν F μ ν = A ν ; μ A μ ; ν F_(mu nu)=A_(nu;mu)-A_(mu;nu)F_{\mu \nu}=A_{\nu ; \mu}-A_{\mu ; \nu}Fμν=Aν;μAμ;ν
These points granted, one can verify that the second of Maxwell's equations is automatically satisfied, and verify also that the first is satisfied if and only if
A α ; μ μ + A μ ; μ α + R α μ A μ = 4 π J α . A α ; μ μ + A μ ; μ α + R α μ A μ = 4 π J α . -A^(alpha;mu)_(mu)+A^(mu)_(;mu)^(alpha)+R^(alpha)_(mu)A^(mu)=4piJ^(alpha).-A^{\alpha ; \mu}{ }_{\mu}+A^{\mu}{ }_{; \mu}^{\alpha}+R^{\alpha}{ }_{\mu} A^{\mu}=4 \pi J^{\alpha} .Aα;μμ+Aμ;μα+RαμAμ=4πJα.
(See §22.4 for fuller discussion and derivation.)
(3) Transport law for Earth's angular-momentum vector. If the Earth were in flat spacetime, like any other isolated body it would parallel-transport its angu-lar-momentum vector S S S\boldsymbol{S}S along the straight world line of its center of mass, u S = 0 u S = 0 grad_(u)S=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{S}=0uS=0 ("conservation of angular momentum"). When translating this transport law into curved spacetime (where the Earth actually resides!), can one ignore curvature coupling? No! Spacetime curvatures due to the moon and sun produce tidal gravitational forces in the Earth; and because the Earth has an equatorial bulge, the tidal forces produce a nonzero net torque about the Earth's center of mass. (In Newtonian language: the piece of bulge nearest the Moon gets pulled with greater force, and hence greater torque, than the piece of bulge farthest from the Moon.) Thus, in curved spacetime one expects a transport law of the form
u S = ( Riemann tensor ) × ( Earth's quadrupole moment ) . u S = (  Riemann tensor  ) × (  Earth's quadrupole moment  ) . grad_(u)S=(" Riemann tensor ")xx(" Earth's quadrupole moment ").\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{S}=(\text { Riemann tensor }) \times(\text { Earth's quadrupole moment }) .uS=( Riemann tensor )×( Earth's quadrupole moment ).
This curvature-coupling torque produces a precession of the Earth's rotation axis through a full circle in the plane of the ecliptic once every 26,000 years ("general precession"; "precession of the equinoxes"; discovered by Hipparchus about 150 в.c.). The precise form of the curvature-coupling term is derived in exercise 16.4.

Exercise 16.4. PRECESSION OF THE EQUINOXES

(a) Show that the transport law for the Earth's intrinsic angular momentum vector S α S α S^(alpha)S^{\alpha}Sα in curved spacetime is
(16.8) D S α D τ = ϵ α β γ δ β μ μ R μ p γ ξ u δ u ν u ζ . (16.8) D S α D τ = ϵ α β γ δ β μ μ R μ p γ ξ u δ u ν u ζ . {:(16.8)(DS^(alpha))/(D tau)=epsilon^(alpha beta gamma delta)_(beta mu mu)R^(mu)_(p gamma xi)u_(delta)u^(nu)u^(zeta).:}\begin{equation*} \frac{D S^{\alpha}}{D \tau}=\epsilon^{\alpha \beta \gamma \delta}{ }_{\beta \mu \mu} R^{\mu}{ }_{p \gamma \xi} u_{\delta} u^{\nu} u^{\zeta} . \tag{16.8} \end{equation*}(16.8)DSαDτ=ϵαβγδβμμRμpγξuδuνuζ.
Here d / d τ = u d / d τ = u d//d tau=ud / d \tau=\boldsymbol{u}d/dτ=u is 4 -velocity along the Earth's world line; t β μ t β μ t_(beta mu)t_{\beta \mu}tβμ is the Earth's "reduced quadrupole moment" (trace-free part of second moment of mass distribution), defined in the Earth's local Lorentz frame by
(16.9) f 0 ^ 0 ^ = f 0 j ^ = 0 , j k ^ = ρ ( x 3 x k ^ 1 3 r ^ 2 δ j k ) d 3 x ^ ; (16.9) f 0 ^ 0 ^ = f 0 j ^ = 0 , j k ^ = ρ x 3 x k ^ 1 3 r ^ 2 δ j k d 3 x ^ ; {:(16.9)f_( hat(0) hat(0))=f_(0 hat(j))=0","quadquad_(j hat(k))=int rho(x^(3)x^( hat(k))-(1)/(3) hat(r)^(2)delta_(jk))d^(3) hat(x);:}\begin{equation*} f_{\hat{0} \hat{0}}=f_{0 \hat{j}}=0, \quad \quad_{j \hat{k}}=\int \rho\left(x^{3} x^{\hat{k}}-\frac{1}{3} \hat{r}^{2} \delta_{j k}\right) d^{3} \hat{x} ; \tag{16.9} \end{equation*}(16.9)f0^0^=f0j^=0,jk^=ρ(x3xk^13r^2δjk)d3x^;
and R μ 2 γ 5 R μ 2 γ 5 R^(mu)_(^(2gamma5))R^{\mu}{ }_{{ }^{2 \gamma 5}}Rμ2γ5 is the Riemann curvature produced at the Earth's location by the moon, sun, and planets. [Hint: Derive this result in the Earth's local Lorentz frame, ignoring the spacetime curvature due to the Earth. (In this essentially Newtonian situation, curvature components R j ^ o ^ k k ^ R j ^ o ^ k k ^ R^( hat(j)_( hat(o)k hat(k)))R^{\hat{j}_{\hat{o} k \hat{k}}}Rj^o^kk^ due to the Earth, sun, moon, and planets superpose linearly; "gravity too weak to be nonlinear"). Integrate up the torque produced about the Earth's center of mass by tidal gravitational forces ("geodesic deviation"):
( acceleration at x 3 ^ , relative to center of mass ( x j ^ = 0 ) , produced by tidal gravitational forces but counterbalanced in part by Earth's internal stresses )  acceleration at  x 3 ^ ,  relative to center of mass  x j ^ = 0 ,  produced by tidal gravitational forces but counterbalanced   in part by Earth's internal stresses  ([" acceleration at "x^( hat(3))","" relative to center of mass "(x^( hat(j))=0)","],[" produced by tidal gravitational forces but counterbalanced "],[" in part by Earth's internal stresses "])\left(\begin{array}{l}\text { acceleration at } x^{\hat{3}}, \text { relative to center of mass }\left(x^{\hat{j}}=0\right), \\ \text { produced by tidal gravitational forces but counterbalanced } \\ \text { in part by Earth's internal stresses }\end{array}\right)( acceleration at x3^, relative to center of mass (xj^=0), produced by tidal gravitational forces but counterbalanced  in part by Earth's internal stresses )
( force per unit volume due to this acceleration, relative to center of mass ) k ^ = mass d 2 x k ^ d t ^ 2 = ρ R k ^ t ^ t ^ ^ x ^ i ^ ^ ; ( torque per unit volume relative to center of mass ) i ^ = ϵ 0 i j k x j ^ ( ρ R k k ^ x 0 x l ^ ; ( total torque about center of mass ) i ^ = [ ϵ 0 i j k x 3 ( ρ R k ^ 0 i 6 x ) ] d 3 x ^ .  force per unit volume due to this   acceleration, relative to center   of mass  k ^ = mass  d 2 x k ^ d t ^ 2 = ρ R k ^ t ^ t ^ ^ x ^ i ^ ^ (  torque per unit volume relative   to center of mass  ) i ^ = ϵ 0 i j k x j ^ ρ R k k ^ x 0 x l ^ (  total torque about center   of mass  ) i ^ = ϵ 0 i j k x 3 ρ R k ^ 0 i 6 x d 3 x ^ . {:[([" force per unit volume due to this "],[" acceleration, relative to center "],[" of mass "])^( hat(k))={:[darr^("mass ")],[(d^(2)x^( hat(k)))/(d hat(t)^(2))=-rhoR^( hat(k)) hat(( hat(t))( hat(hat(t)))( hat(x))( hat(i)))"; "]:}],[((" torque per unit volume relative ")/(" to center of mass "))_( hat(i))=epsilon_(0ijk)x^( hat(j))(-rhoR^(k)( hat(k))x_(0)x^( hat(l)):}"; "],[((" total torque about center ")/(" of mass "))_( hat(i))=int[epsilon_(0ijk)x^(3)(-rhoR^( hat(k))_(0i6)x^(ℓ))]d^(3) hat(x).]:}\begin{aligned} & \left(\begin{array}{l} \text { force per unit volume due to this } \\ \text { acceleration, relative to center } \\ \text { of mass } \end{array}\right)^{\hat{k}}=\begin{array}{l} \downarrow^{\text {mass }} \\ \frac{d^{2} x^{\hat{k}}}{d \hat{t}^{2}}=-\rho R^{\hat{k}} \hat{\hat{t} \hat{\hat{t}} \hat{x} \hat{i}} \text {; } \end{array} \\ & \binom{\text { torque per unit volume relative }}{\text { to center of mass }}_{\hat{i}}=\epsilon_{0 i j k} x^{\hat{j}}\left(-\rho R^{k} \hat{k} x_{0} x^{\hat{l}}\right. \text {; } \\ & \binom{\text { total torque about center }}{\text { of mass }}_{\hat{i}}=\int\left[\epsilon_{0 i j k} x^{3}\left(-\rho R^{\hat{k}}{ }_{0 i 6} x^{\ell}\right)\right] d^{3} \hat{x} . \end{aligned}( force per unit volume due to this  acceleration, relative to center  of mass )k^=mass d2xk^dt^2=ρRk^t^t^^x^i^^( torque per unit volume relative  to center of mass )i^=ϵ0ijkxj^(ρRkk^x0xl^( total torque about center  of mass )i^=[ϵ0ijkx3(ρRk^0i6x)]d3x^.
Put this expression into a form involving f j j ^ f j j ^ f_(j)_( hat(j)){f_{j}}_{\hat{j}}fjj^, equate it to d S i / d τ d S i / d τ dS_(i)//d taud S_{i} / d \taudSi/dτ, and then reexpress it in frame-independent, component notation. The result should be equation (16.8).]
(b) Rewrite equation (16.8) in the Earth's local Lorentz frame, using the equation
R ^ k ^ 0 ^ 0 ^ = 2 Φ / x j ^ x k ^ R ^ k ^ 0 ^ 0 ^ = 2 Φ / x j ^ x k ^ R_( hat(del) hat(k) hat(0))^( hat(0))=del^(2)Phi//delx^( hat(j))delx^( hat(k))R_{\hat{\partial} \hat{k} \hat{0}}^{\hat{0}}=\partial^{2} \Phi / \partial x^{\hat{j}} \partial x^{\hat{k}}R^k^0^0^=2Φ/xj^xk^
for the components of Riemann in terms of the Newtonian gravitational potential. (Newtonian approximation to Einstein theory. Track-2 readers have met this equation in Chapter 12; track-one readers will meet it in $17.4.)
(c) Calculate d S j ^ / d d ^ d S j ^ / d d ^ dS^( hat(j))//d hat(d)d S^{\hat{j}} / d \hat{d}dSj^/dd^ using Newton's theory of gravity from the beginning. The answer should be identical to that obtained in part (b) using Einstein's theory.
(d) Idealizing the moon and sun as point masses, calculate the long-term effect of the spacetime curvatures that they produce upon the Earth's rotation axis. Use the result of part (b), together with moderately accurate numerical values for the relevant solar-system parameters. [Answer: The Earth's rotation axis precesses relative to the axes of its local Lorentz frame ("precession of the equinoxes"; "general precession"); the precession period is 26,000 years. The details of the calculation will be found in any textbook on celestial mechanics.]

§16.4. THE RODS AND CLOCKS USED TO MEASURE SPACE AND TIME INTERVALS

Turn attention now from the laws of physics in the presence of gravity to the nature of the rods and clocks that must be used for measuring the length and time intervals appearing in those laws.
One need not-and indeed must not!-postulate that proper length s s sss is measured by a certain type of rod (e.g., platinum meter stick), or that proper time τ τ tau\tauτ is measured by a certain type of clock (e.g., hydrogen-maser clock). Rather, one must ask the laws of physics themselves what types of rods and clocks will do the job. Put differently, one defines an "ideal" rod or clock to be one which measures proper length as given by d s = ( g α β d x α d x β ) 1 / 2 d s = g α β d x α d x β 1 / 2 ds=(g_(alpha beta)dx^(alpha)dx^(beta))^(1//2)d s=\left(g_{\alpha \beta} d x^{\alpha} d x^{\beta}\right)^{1 / 2}ds=(gαβdxαdxβ)1/2 or proper time as given by d τ = d τ = d tau=d \tau=dτ= ( g α β d x α d x β ) 1 / 2 g α β d x α d x β 1 / 2 (-g_(alpha beta)dx^(alpha)dx^(beta))^(1//2)\left(-g_{\alpha \beta} d x^{\alpha} d x^{\beta}\right)^{1 / 2}(gαβdxαdxβ)1/2 (the kind of clock to which one was led by physical arguments in §1.5). One must then determine the accuracy to which a given rod or clock is ideal under given circumstances by using the laws of physics to analyze its behavior.
As an obvious example, consider a pendulum clock. If it is placed at rest on the Earth's surface, if it is tiny enough that redshift effects from one end to the other and time dilation effects due to its swinging velocity are negligible, and if the accuracy one demands is small enough that time variations in the local gravitational acceleration due to Earth tides can be ignored, then the laws of physics report (Box 16.2) that the pendulum clock is "ideal." However, in any other context (e.g., on a rocket journey to the moon), a pendulum clock should be far from ideal. Wildly changing accelerations, or no acceleration at all, will make it worthless!
Of greater interest are atomic and nuclear clocks of various sorts. Such a clock is analyzed most easily if it is freely falling. One can then study it in its local Lorentz rest frame, using the standard equations of quantum theory; and, of course, one will find that it measures proper time to within the precision ( Δ t / t 10 9 Δ t / t 10 9 (Delta t//t∼10^(-9):}\left(\Delta t / t \sim 10^{-9}\right.(Δt/t109 to 10 14 ) 10 14 {:10^(-14))\left.10^{-14}\right)1014) of the technology used in its construction. However, one rarely permits his atomic clock to fall freely. (The impact with the Earth's surface can be expensive!) Nevertheless, even when accelerated at " 1 g " = 980 cm / sec 2 = 980 cm / sec 2 =980cm//sec^(2)=980 \mathrm{~cm} / \mathrm{sec}^{2}=980 cm/sec2 on the Earth's surface, and even when accelerated at " 2 g " in an airliner trying to avoid a midair collision (Box 16.3), an atomic clock-if built solidly-will still measure proper time d τ = d τ = d tau=d \tau=dτ= ( g α β d x α d x β ) 1 / 2 g α β d x α d x β 1 / 2 (-g_(alpha beta)dx^(alpha)dx^(beta))^(1//2)\left(-g_{\alpha \beta} d x^{\alpha} d x^{\beta}\right)^{1 / 2}(gαβdxαdxβ)1/2 along its world line to nearly the same accuracy as if it were freely falling. To discover this one can perform an experiment. Alternatively, one can analyze the clock in its own "proper reference frame" (§13.6), with Fermi-Walkertransported basis vectors, using the standard local Lorentz laws of quantum mechanics as adapted to accelerated frames (local Lorentz laws plus an "inertial force," which can be treated as due to a potential with a uniform gradient.
Of course, any clock has a "breaking point," beyond which it will cease to function properly (Box 16.3). But that breaking point depends entirely on the construction of the clock-and not at all on any "universal influence of acceleration on the march of time." Velocity produces a universal time dilation; acceleration does not.
The aging of the human body is governed by the same electromagnetic and quantum-mechanical laws as govern the periodicities and level transitions in atoms and molecules. Consequently, aging, like atomic processes, is tied to proper time
"Ideal" rods and clocks defined
How ideal are real clocks?
(1) pendulum clocks
(2) atomic clocks
(3) human clocks

Box 16.2 PROOF THAT A PENDULUM CLOCK AT REST ON THE EARTH'S SURFACE IS IDEAL

That is, a proof that it measures the interval d τ = ( g α β d x α d x β ) 1 / 2 d τ = g α β d x α d x β 1 / 2 d tau=(-g_(alpha beta)dx^(alpha)dx^(beta))^(1//2)d \tau=\left(-g_{\alpha \beta} d x^{\alpha} d x^{\beta}\right)^{1 / 2}dτ=(gαβdxαdxβ)1/2.

A. Constraint on the Pendulum

It must be so small that it cannot couple to the spacetime curvature-i.e., so small that the Earth's gravitational field looks uniform in its neighborhood-and that the velocity of its ball is totally negligible compared to the speed of light.

B. Coordinate System and Metric

(1) General coordinate system: because the Earth's field is nearly Newtonian, one can introduce the coordinates of "linearized theory" ( $ 18.4 $ 18.4 $18.4\$ 18.4$18.4; one must take this on faith until one reaches that point) in which
d s 2 = ( 1 + 2 Φ ) d t 2 + ( 1 2 Φ ) ( d x 2 + d y 2 + d z 2 ) d s 2 = ( 1 + 2 Φ ) d t 2 + ( 1 2 Φ ) d x 2 + d y 2 + d z 2 ds^(2)=-(1+2Phi)dt^('2)+(1-2Phi)(dx^('2)+dy^('2)+dz^('2))d s^{2}=-(1+2 \Phi) d t^{\prime 2}+(1-2 \Phi)\left(d x^{\prime 2}+d y^{\prime 2}+d z^{\prime 2}\right)ds2=(1+2Φ)dt2+(12Φ)(dx2+dy2+dz2),
where Φ Φ Phi\PhiΦ is the Newtonian potential.

(2) Put the origin of coordinates at the pendulum's equilibrium position, and orient the x , z x , z x^('),z^(')x^{\prime}, z^{\prime}x,z-plane so the pendulum swings in it.
(3) Renormalize the coordinates so they measure proper length and proper time at the equilibrium position
t = [ 1 + 2 Φ ( 0 ) ] 1 / 2 t , x j = [ 1 2 Φ ( 0 ) ] 1 / 2 x j t = [ 1 + 2 Φ ( 0 ) ] 1 / 2 t , x j = [ 1 2 Φ ( 0 ) ] 1 / 2 x j t=[1+2Phi(0)]^(1//2)t^('),quadx^(j)=[1-2Phi(0)]^(1//2)x^(j^('))t=[1+2 \Phi(0)]^{1 / 2} t^{\prime}, \quad x^{j}=[1-2 \Phi(0)]^{1 / 2} x^{j^{\prime}}t=[1+2Φ(0)]1/2t,xj=[12Φ(0)]1/2xj
Then near the pendulum (inhomogeneities in the field neglected!)
(1) Φ = Φ ( 0 ) + g z , g = "acceleration of gravity," (2) d s 2 = ( 1 + 2 g z ) d t 2 + ( 1 2 g z ) ( d x 2 + d y 2 + d z 2 ) . (1) Φ = Φ ( 0 ) + g z , g =  "acceleration of gravity,"  (2) d s 2 = ( 1 + 2 g z ) d t 2 + ( 1 2 g z ) d x 2 + d y 2 + d z 2 . {:[(1)Phi=Phi(0)+gz","quad g=" "acceleration of gravity," "],[(2)ds^(2)=-(1+2gz)dt^(2)+(1-2gz)(dx^(2)+dy^(2)+dz^(2)).]:}\begin{align*} \Phi & =\Phi(0)+g z, \quad g=\text { "acceleration of gravity," } \tag{1}\\ d s^{2} & =-(1+2 g z) d t^{2}+(1-2 g z)\left(d x^{2}+d y^{2}+d z^{2}\right) . \tag{2} \end{align*}(1)Φ=Φ(0)+gz,g= "acceleration of gravity," (2)ds2=(1+2gz)dt2+(12gz)(dx2+dy2+dz2).

C. Analysis of Pendulum Motion

(1) Put the total mass m m mmm of the pendulum in its ball (negligible mass in its rod). Let its rod have proper length l l lll.
(2) Calculate the 4-acceleration a = u u a = u u a=grad_(u)u\boldsymbol{a}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}a=uu of the pendulum's ball in terms of d 2 x α / d t 2 d 2 x α / d t 2 d^(2)x^(alpha)//dt^(2)d^{2} x^{\alpha} / d t^{2}d2xα/dt2, using the velocity condition v 1 v 1 v⋘1v \lll 1v1 and d t / d τ 1 d t / d τ 1 dt//d tau~~1d t / d \tau \approx 1dt/dτ1 :
a x = d 2 x / d τ 2 + Γ x 00 ( d t / d τ ) 2 = d 2 x / d t 2 + Γ x 00 = d 2 x / d t 2 , (3) a z = d 2 z / d τ 2 + Γ z 00 ( d t / d τ ) 2 = d 2 z / d t 2 + Γ z 00 = d 2 z / d t 2 + g . a x = d 2 x / d τ 2 + Γ x 00 ( d t / d τ ) 2 = d 2 x / d t 2 + Γ x 00 = d 2 x / d t 2 , (3) a z = d 2 z / d τ 2 + Γ z 00 ( d t / d τ ) 2 = d 2 z / d t 2 + Γ z 00 = d 2 z / d t 2 + g . {:[a^(x)=d^(2)x//dtau^(2)+Gamma^(x)_(00)(dt//d tau)^(2)=d^(2)x//dt^(2)+Gamma^(x)_(00)=d^(2)x//dt^(2)","],[(3)a^(z)=d^(2)z//dtau^(2)+Gamma^(z)_(00)(dt//d tau)^(2)=d^(2)z//dt^(2)+Gamma^(z)_(00)=d^(2)z//dt^(2)+g.]:}\begin{align*} & a^{x}=d^{2} x / d \tau^{2}+\Gamma^{x}{ }_{00}(d t / d \tau)^{2}=d^{2} x / d t^{2}+\Gamma^{x}{ }_{00}=d^{2} x / d t^{2}, \\ & a^{z}=d^{2} z / d \tau^{2}+\Gamma^{z}{ }_{00}(d t / d \tau)^{2}=d^{2} z / d t^{2}+\Gamma^{z}{ }_{00}=d^{2} z / d t^{2}+g . \tag{3} \end{align*}ax=d2x/dτ2+Γx00(dt/dτ)2=d2x/dt2+Γx00=d2x/dt2,(3)az=d2z/dτ2+Γz00(dt/dτ)2=d2z/dt2+Γz00=d2z/dt2+g.
(3) This 4 -acceleration must be produced by the forces in the rod, and must be directed up the rod so that (for x l x l x≪lx \ll lxl so g d 2 z / d t 2 g d 2 z / d t 2 g≫d^(2)z//dt^(2)g \gg d^{2} z / d t^{2}gd2z/dt2 )
(4) d 2 x / d t 2 = a x = ( x / l ) a z = ( g / l ) x (4) d 2 x / d t 2 = a x = ( x / l ) a z = ( g / l ) x {:(4)d^(2)x//dt^(2)=a^(x)=-(x//l)a^(z)=-(g//l)x:}\begin{equation*} d^{2} x / d t^{2}=a^{x}=-(x / l) a^{z}=-(g / l) x \tag{4} \end{equation*}(4)d2x/dt2=ax=(x/l)az=(g/l)x
(4) Solve this differential equation to obtain
(5) x = x 0 cos ( t g / l ) . (5) x = x 0 cos ( t g / l ) . {:(5)x=x_(0)cos(tsqrt(g//l)).:}\begin{equation*} x=x_{0} \cos (t \sqrt{g / l}) . \tag{5} \end{equation*}(5)x=x0cos(tg/l).
(5) Thus conclude that the pendulum is periodic in t t ttt, which is proper time at the ball's equilibrium position (see equation 2). This means that the pendulum is an ideal clock when it is at rest on the Earth's surface.
Note: The above analysis ignores the Earth's rotation; for an alternative analysis including rotation, one can perform a similar calculation at the origin of the pendulum's "proper reference frame" [ § 13.6 [ § 13.6 [§13.6[\S 13.6[§13.6; line element (13.71)]. The answer is the same; but now " g g ggg " is a superposition of the "gravitational acceleration," and the "centrifugal acceleration produced by Earth's rotation."

Box 16.3 RESPONSE OF CLOCKS TO ACCELERATION AND TO TIDAL GRAVITATIONAL FORCES

Consider an atomic clock with frequency stabilized by some atomic or molecular process-for example, fixed by the "umbrella vibrations" of ammonia molecules [see Feynman et. al. (1964)]. When subjected to sufficiently strong accelerations or tidal forces, such a clock will cease to measure proper time with its normal precision. Two types of effects could lead to such departures from "ideality":
A. Influence of the acceleration or tidal force on the atomic process that provides the frequency stability. Example: If tidal forces are significant over distances of a few angstroms (e.g., near a spacetime "singularity" terminating gravitational collapse), then they can and will deform an ammonia molecule and destroy the regularity of its umbrella vibrations, thereby making useless any ammonia atomic clock, no matter how constructed. Similarly, if an ammonia molecule is subjected to accelerations of magnitude comparable to its internal atomic accelerations ( a 10 12 " g " 10 15 cm / a 10 12 " g " 10 15 cm / (a∼10^(12")(g)"∼10^(15)(cm)//:}\left(a \sim 10^{12 "} \mathrm{~g} " \sim 10^{15} \mathrm{~cm} /\right.(a1012" g"1015 cm/ sec 2 sec 2 sec^(2)\mathrm{sec}^{2}sec2 ), which change in times of the order of the "umbrella" vibration period, then it must cease to vibrate regularly, and any clock based on its vibrations must fail. Such limits of principle on the ideality of a clock will vary from one atomic process to another. However, they are far from being a limiting factor on clock construction in 1973. Much more important today is:
B. Influence of the acceleration or tidal force on the macroscopic structure of the clock-a structure dictated by current technology. The crystal oscillator,
which produces the periodic signal output, must be locked to the regulating atomic process in some way. The lock will be disturbed by moderate accelerations. The toughest task for the manufacturer of aircraft clocks is to guarantee that precise locking will be maintained, even when the aircraft is maneuvering desperately to avoid collision with another aircraft or with a missile. In 1972 a solidly built rubidium clock will maintain its lock, with no apparent degradation of stability
[ Δ t / t 10 12 ( 1 sec / t ) 1 / 2 Δ t / t 10 12 ( 1 sec / t ) 1 / 2 [Delta t//t∼10^(-12)(1sec//t)^(1//2):}\left[\Delta t / t \sim 10^{-12}(1 \mathrm{sec} / t)^{1 / 2}\right.[Δt/t1012(1sec/t)1/2 for 1 sec t 10 3 sec ] 1 sec t 10 3 sec {:1sec <= t <= 10^(3)sec]\left.1 \mathrm{sec} \leqq t \leqq 10^{3} \mathrm{sec}\right]1sect103sec] under steady-state accelerations up to 50 " g g ggg " or more. But, because of the finite bandwidth of the lock loop (typically Δ v 20 Δ v 20 Delta v∼20\Delta v \sim 20Δv20 to 50 Hz ), sudden changes in acceleration will temporarily break the lock, degrading the clock stability to that of the unlocked crystal oscillator-for which an acceleration a a aaa produces a change in frequency of about ( a / 1 a / 1 (a//1:}\left(a / 1\right.(a/1 " g g ggg ") × 10 9 × 10 9 xx10^(-9)\times 10^{-9}×109. But the lock to the rubidium standard is restored quickly ( δ t 1 / Δ ν δ t 1 / Δ ν delta t∼1//Delta nu\delta t \sim 1 / \Delta \nuδt1/Δν ), bringing the clock back to its normal highly stable performance.*
Tidal forces are so small in the solar system that the clock manufacturer can ignore them. However, a 1973 atomic clock, subjected to the tidal accelerations near a spacetime singularity, should break the "lock" to its atomic process long before the tidal forces become strong enough to influence the atomic process itself.
Ideal rods and clocks constructed from geodesic world lines
as governed by the metric-though, of course, it is also tied to other things, such as cigarette smoking.
In principle, one can build ideal rods and clocks from the geodesic world lines of freely falling test particles and photons. (See Box 16.4.) In other words, spacetime has its own rods and clocks built into itself, even when matter and nongravitational fields are absent!

Box 16.4 IDEAL RODS AND CLOCKS BUILT FROM GEODESIC WORLD LINES*

The Standard Interval. A specific timelike inter-val-the interval between two particular neighboring events a a a\mathscr{a}a and B B B\mathscr{B}B-is chosen as the standard interval, and is assigned unit length. It is used to calibrate a huge set of geodesic clocks that pass through a a aaa.
Each geodesic clock is constructed and calibrated as follows:
(1) A timelike geodesic C C C C CC\mathscr{C C}CC (path of freely falling particle) passes through a a aaa.
(2) A neighboring world line, everywhere parallel to A C A C AC\mathscr{A C}AC (and thus not a geodesic), is constructed by the method of Schild's ladder (Box 10.2), which relies only on geodesics.
(3) Light rays (null geodesics) bounce back and forth between these parallel world lines; each round trip constitutes one "tick."
(4) The proper time lapse, τ 0 τ 0 tau_(0)\tau_{0}τ0, between ticks is related to the interval C G B C G B CGB\mathscr{C G B}CGB by
1 ( Q B ) 2 = ( N 1 τ 0 ) ( N 2 τ 0 ) , 1 ( Q B ) 2 = N 1 τ 0 N 2 τ 0 , -1-=(QB)^(2)=-(N_(1)tau_(0))(N_(2)tau_(0)),-1 \equiv(\mathscr{Q} \mathscr{B})^{2}=-\left(N_{1} \tau_{0}\right)\left(N_{2} \tau_{0}\right),1(QB)2=(N1τ0)(N2τ0),
where N 1 N 1 N_(1)N_{1}N1 and N 2 N 2 N_(2)N_{2}N2 are the number of ticks between the events shown in the diagrams. [Proof: see diagram at right.]
Spacetime is filled with such geodesic clocks. Those that pass through C C C\mathscr{C}C are calibrated as above against the standard interval a B a B aBa \mathscr{B}aB, and are used subsequently to calibrate all other clocks they meet.

In local Lorentz rest frame of geodesic clock:
( N 1 τ 0 ) ( N 2 τ 0 ) = ( t x ) ( t + x ) = t 2 x 2 = ( a G ) 2 N 1 τ 0 N 2 τ 0 = ( t x ) ( t + x ) = t 2 x 2 = ( a G ) 2 {:[(N_(1)tau_(0))(N_(2)tau_(0))=(t-x)(t+x)],[=t^(2)-x^(2)=-(aG)^(2)]:}\begin{aligned} \left(N_{1} \tau_{0}\right)\left(N_{2} \tau_{0}\right) & =(t-x)(t+x) \\ & =t^{2}-x^{2}=-(a \mathscr{G})^{2} \end{aligned}(N1τ0)(N2τ0)=(tx)(t+x)=t2x2=(aG)2

Box 16.4 (continued)

Any interval P Q P Q PQ\mathscr{P Q}PQ along the world line of a geodesic clock can be measured by the same method as was used in calibration. The interval P Q P Q PQ\mathscr{P Q}PQ can be timelike, spacelike, or null; its squared length in all three cases will be
( P Q ) 2 = ( N 3 τ 0 ) ( N 4 τ 0 ) ( P Q ) 2 = N 3 τ 0 N 4 τ 0 (PQ)^(2)=-(N_(3)tau_(0))(N_(4)tau_(0))(\mathscr{P Q})^{2}=-\left(N_{3} \tau_{0}\right)\left(N_{4} \tau_{0}\right)(PQ)2=(N3τ0)(N4τ0)
To achieve a precision of measurement good to one part in N N NNN, where N N NNN is some large number, take two precautions:

(1) Demand that the intervals Q B Q B QB\mathscr{Q} \mathscr{B}QB and P Q P Q PQ\mathscr{P} \mathscr{Q}PQ be sufficiently small compared to the scale of curvature of spacetime; or specifically,
R ( A B ) ( Q G B ) 2 1 / N R ( A B ) ( Q G B ) 2 1 / N R^((AB))(QGB)^(2)≪1//NR^{(A B)}(\mathscr{Q G B})^{2} \ll 1 / NR(AB)(QGB)21/N
and
R ( P Q ) ( P Q 2 ) 2 1 / N , R ( P Q ) ( P Q 2 ) 2 1 / N , R^((PQ))(PQ2)^(2)≪1//N,R^{(P Q)}(\mathscr{P Q} \mathscr{2})^{2} \ll 1 / N,R(PQ)(PQ2)21/N,
where R ( A B ) R ( A B ) R^((AB))R^{(A B)}R(AB) and R ( P Q ) R ( P Q ) R^((PQ))R^{(P Q)}R(PQ) are the largest relevant components of the curvature tensor in the two regions in question.
(2) Demand that the time scale, τ 0 τ 0 tau_(0)\tau_{0}τ0, of the geodesic clocks employed be small compared to C B C B CB\mathscr{C B}CB and P Q P Q PQ\mathscr{P} \mathcal{Q}PQ individually; thus,
τ 0 A Q / N , τ 0 P Q / N . τ 0 A Q / N , τ 0 P Q / N . {:[tau_(0)≪AQ//N","],[tau_(0)≪PQ//N.]:}\begin{aligned} & \tau_{0} \ll \mathscr{A Q} / N, \\ & \tau_{0} \ll \mathscr{P Q} / N . \end{aligned}τ0AQ/N,τ0PQ/N.
The Einstein principle that spacetime is described by Riemannian geometry exposes itself to destruction by a "thousand" tests. Thus, from the fiducial interval, A B A B AB\mathscr{A B}AB, to the interval under measurement, P Q P Q PQ\mathscr{P Q}PQ, there are a "score" of routes of intercomparison, all of which must give the same value for the ratio P Q / Q B P Q / Q B PQ//QB\mathscr{P} \mathscr{Q} / \mathscr{Q} \mathscr{B}PQ/QB. Moreover, one can easily select out "fifty" intervals P Q P Q PQ\mathscr{P} \mathscr{Q}PQ to which the same kind of test can be applied. Such tests are not all items for the future.
Some 5 × 10 9 5 × 10 9 5xx10^(9)5 \times 10^{9}5×109 years ago, electrons arrived by different routes at a common location, a given atom of iron in the core of the earth. This iron atom does not collapse. The Pauli principle of
exclusion keeps the electrons from all falling into the K-orbit. The Pauli principle would not apply if the electrons were not identical or nearly so. From this circumstance it would appear possible to draw an important conclusion (Marzke and Wheeler). With each electron is associated a standard length, its Compton wavelength, / m c / m c ℏ//mc\hbar / m c/mc. If these lengths had started different, or changed by different amounts along the different routes, and if the resulting difference in properties were as great as one part in
( 5 × 10 9 yr ) × ( 3 × 10 7 sec / yr ) × ( 5 × 10 18 rev / sec ) 10 36 , 5 × 10 9 yr × ( 3 × 10 7 sec / yr × 5 × 10 18 rev / sec 10 36 , {:[∼(5xx10^(9)yr)xx(3{: xx10^(7)sec//yr)],[ xx(5xx10^(18)rev//sec)∼10^(36)","]:}\begin{aligned} \sim\left(5 \times 10^{9} \mathrm{yr}\right) \times(3 & \left.\times 10^{7} \mathrm{sec} / \mathrm{yr}\right) \\ & \times\left(5 \times 10^{18} \mathrm{rev} / \mathrm{sec}\right) \sim 10^{36}, \end{aligned}(5×109yr)×(3×107sec/yr)×(5×1018rev/sec)1036,
by now this difference would have shown up, the varied electrons would have fallen into the Korbit, and the earth would have collapsed, contrary to observation.
The Marzke-Wheeler construction expresses an arbitrary small interval P Q P Q PQ\mathscr{P Q}PQ, anywhere in spacetime, in terms of the fiducial interval C O B C O B COB\mathscr{C O B}COB, an interval which itself may be taken for definiteness to be the "geometrodynamic standard centimeter" of §1.5. This construction thus gives a vivid meaning to the idea of Riemannian geometry.
The M-W construction makes no appeal what-
soever to rods and clocks of atomic constitution. This circumstance is significant for the following reasons. The length of the usual platinum meter stick is some multiple, N 1 ( 2 / m e 2 ) N 1 2 / m e 2 N_(1)(ℏ^(2)//me^(2))N_{1}\left(\hbar^{2} / m e^{2}\right)N1(2/me2), of the Bohr atomic radius. Similarly, the wavelength of the Kr 86 Kr 86 Kr^(86)\mathrm{Kr}^{86}Kr86 line is some multiple, N 2 ( c / e 2 ) ( 2 / m e 2 ) N 2 c / e 2 2 / m e 2 N_(2)(ℏc//e^(2))(ℏ^(2)//me^(2))N_{2}\left(\hbar c / e^{2}\right)\left(\hbar^{2} / m e^{2}\right)N2(c/e2)(2/me2), of a second basic length that depends on the atomic constants in quite a different way. Thus, if there is any change with time in the dimensionless ratio c / e 2 = 137.038 c / e 2 = 137.038 ℏc//e^(2)=137.038\hbar c / e^{2}=137.038c/e2=137.038, one or the other or both of these atomic standards of length must get out of kilter with the geometrodynamic standard centimeter. In this case, general relativity says, "Stick to the geometrodynamic standard centimeter."
Hermann Weyl at first thought that one could carry out the comparison of lengths by light rays alone, but H . A. Lorentz pointed out that one can dispense with the geodesics neither of test particles nor of light rays in the measurement process, the construction for which, however, neither Weyl nor Lorentz supplied [literature in Marzke and Wheeler (1964)]. Ehlers, Pirani, and Schild (1972) have given a deeper analysis of the separate parts played in the measurement process by the affine connection, by the conformal part of the metric, and by the full metric.

§16.5. THE MEASUREMENT OF THE GRAVITATIONAL FIELD

"I know how to measure the electromagnetic field using test charges; what is the analogous procedure for measuring the gravitational field?" This question has, at the same time, many answers and none.
It has no answers because nowhere has a precise definition of the term "gravitational field" been given-nor will one be given. Many different mathematical entities are associated with gravitation: the metric, the Riemann curvature tensor, the Ricci curvature tensor, the curvature scalar, the covariant derivative, the connection coefficients, etc. Each of these plays an important role in gravitation theory, and none is so much more central than the others that it deserves the name "gravitational field." Thus it is that throughout this book the terms "gravitational field" and "gravity" refer in a vague, collective sort of way to all of these entities. Another, equivalent term used for them is the "geometry of spacetime."
The many faces of gravity, and how one measures them
To "measure the gravitational field," then, means to "explore experimentally various properties of the spacetime geometry." One makes different kinds of measurements, depending on which geometric property of spacetime one is interested in. However, all such measurements must involve a scrutiny of the effects of the spacetime geometry (i.e., of gravity) on particles, on matter, or on nongravitational fields.
For example, to "measure" the metric near a given event, one typically lays out a latticework of rods and clocks (local orthonormal frame, small enough that curvature effects are negligible), and uses it to determine the interval between neighboring events. To measure the Riemann curvature tensor near an event, one typically studies the geodesic deviation (relative accelerations) that curvature produces between the world lines of a variety of neighboring test particles; alternatively, one makes measurements with a "gravity gradiometer" (Box 16.5) if the curvature is static or slowly varying; or with a gravitational wave antenna (Chapter 37) if the curvature fluctuates rapidly. To study the large-scale curvature of spacetime, one examines large-scale effects of gravity, such as the orbits of planets and satellites, or the bending of light by the sun's gravitational field.
But whatever aspect of gravity one measures, and however one measures it, one is studying the geometry of spacetime.

EXERCISE

Exercise 16.5. GRAVITY GRADIOMETER

The gravity gradiometer of Box 16.5 moves through curved spacetime along an accelerated world line. Calculate the amplitude and phase of oscillation of one arm of the gradiometer relative to the other. [Hint: Perform the calculation in the gradiometer's "proper reference frame" ($13.6), with Fermi-Walker-transported basis vectors. Use, as the equation for the relative angular acceleration of the two arms,
2 m 2 ( α ¨ + α ˙ / τ 0 + ω 0 2 α ) = ( Driving torque produced by Riemann curvature ) , 2 m 2 α ¨ + α ˙ / τ 0 + ω 0 2 α = (  Driving torque produced by   Riemann curvature  ) , 2mℓ^(2)((alpha^(¨))+(alpha^(˙))//tau_(0)+omega_(0)^(2)alpha)=((" Driving torque produced by ")/(" Riemann curvature ")),2 m \ell^{2}\left(\ddot{\alpha}+\dot{\alpha} / \tau_{0}+\omega_{0}^{2} \alpha\right)=\binom{\text { Driving torque produced by }}{\text { Riemann curvature }},2m2(α¨+α˙/τ0+ω02α)=( Driving torque produced by  Riemann curvature ),
where
2 m 2 = (moment of inertia of one arm) , α = (angular displacement of one arm from equilibrium) , π 2 + 2 α = (angular separation of the two arms), 2 m 2 ω 0 2 = (torsional spring constant), ω 0 = (angular frequency of free vibrations), τ 0 = (decay time for free vibrations to damp out due to internal frictional forces) . 2 m 2 =  (moment of inertia of one arm)  , α =  (angular displacement of one arm from equilibrium)  , π 2 + 2 α =  (angular separation of the two arms),  2 m 2 ω 0 2 =  (torsional spring constant),  ω 0 =  (angular frequency of free vibrations),  τ 0 =  (decay time for free vibrations to damp out due to   internal frictional forces)  . {:[2mℓ^(2)=" (moment of inertia of one arm) "","],[alpha=" (angular displacement of one arm from equilibrium) "","],[(pi)/(2)+2alpha=" (angular separation of the two arms), "],[2mℓ^(2)omega_(0)^(2)=" (torsional spring constant), "],[omega_(0)=" (angular frequency of free vibrations), "],[tau_(0)=" (decay time for free vibrations to damp out due to "],[" internal frictional forces) ".]:}\begin{aligned} 2 m \ell^{2} & =\text { (moment of inertia of one arm) }, \\ \alpha & =\text { (angular displacement of one arm from equilibrium) }, \\ \frac{\pi}{2}+2 \alpha & =\text { (angular separation of the two arms), } \\ 2 m \ell^{2} \omega_{0}^{2} & =\text { (torsional spring constant), } \\ \omega_{0} & =\text { (angular frequency of free vibrations), } \\ \tau_{0} & =\text { (decay time for free vibrations to damp out due to } \\ & \text { internal frictional forces) } . \end{aligned}2m2= (moment of inertia of one arm) ,α= (angular displacement of one arm from equilibrium) ,π2+2α= (angular separation of the two arms), 2m2ω02= (torsional spring constant), ω0= (angular frequency of free vibrations), τ0= (decay time for free vibrations to damp out due to  internal frictional forces) .
If ξ ξ xi\xiξ is the vector from the center of mass of the gradiometer to mass 1 , then one has
( curvature-produced acceleration of mass 1 relative to center of gradiometer ) k ^ = ( D 2 ξ k ^ d τ 2 ) geodesesic deviation = R k 000 ξ ;  curvature-produced   acceleration of mass  1  relative to center of   gradiometer  k ^ = D 2 ξ k ^ d τ 2  geodesesic   deviation  = R k 000 ξ ; ([" curvature-produced "],[" acceleration of mass "1],[" relative to center of "],[" gradiometer "])_( hat(k))=((D^(2)xi_( hat(k)))/(dtau^(2)))_({:[" geodesesic "],[" deviation "]:})=-R_(k 000 xi);\left(\begin{array}{l} \text { curvature-produced } \\ \text { acceleration of mass } 1 \\ \text { relative to center of } \\ \text { gradiometer } \end{array}\right)_{\hat{k}}=\left(\frac{D^{2} \xi_{\hat{k}}}{d \tau^{2}}\right)_{\substack{\text { geodesesic } \\ \text { deviation }}}=-R_{k 000 \xi} ;( curvature-produced  acceleration of mass 1 relative to center of  gradiometer )k^=(D2ξk^dτ2) geodesesic  deviation =Rk000ξ;

Box 16.5 GRAVITY GRADIOMETER FOR MEASURING THE RIEMANN CURVATURE OF SPACETIME

This gravity gradiometer was designed and built by Robert M. Forward and his colleagues at Hughes Research Laboratories, Malibu, California. It measures the Riemann curvature of spacetime produced by nearby masses. By flying a more advanced version of such a gradiometer in an airplane above the Earth's surface, one should be able to measure subsurface mass variations due to varying geological structure. In an Earth-orbiting satellite, such a gradiometer could measure the gravitational multipole moments of the Earth. Technical details of the gradiometer are spelled out in the papers of Forward (1972), and Bell, Forward, and Williams (1970). The principles of its operation are outlined below.
The gradiometer consists of two orthogonal arms with masses m m mmm on their ends, connected at their centers by a torsional spring. When the arms are twisted out of orthogonal alignment, they oscillate. A piezoelectric strain transducer is used to measure the oscillation amplitude.

Box 16.5 (continued)

When placed near an external mass, M M MMM, the gradiometer experiences a torque: because of the gradient in the gravitational field of M M MMM (i.e., because of the spacetime curvature produced by M M MMM ), the Newtonian forces F 1 F 1 F_(1)F_{1}F1 and F 2 F 2 F_(2)F_{2}F2 are greater than F 3 F 3 F_(3)F_{3}F3 and F 4 F 4 F_(4)F_{4}F4; so a net torque pulls masses 1 and 2 toward each other, and 3 and 4 toward each other. [Note: the forces F 1 , F 2 , F 3 , F 4 F 1 , F 2 , F 3 , F 4 F_(1),F_(2),F_(3),F_(4)F_{1}, F_{2}, F_{3}, F_{4}F1,F2,F3,F4 depend on whether the gradiometer is in free fall (geodesic motion; u u = 0 u u = 0 grad_(u)u=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=0uu=0 ) or is moving on an accelerated world line. But the net torque is unaffected by acceleration; acceleration produces equal Newtonian
forces on all four masses, with zero net torque.]
When in operation the gradiometer rotates with angular velocity ω ω omega\omegaω about its center. As it rotates, the torques on its arms oscillate:
at ω t = 0 ω t = 0 omega t=0\omega t=0ωt=0 net torque pushes 1 and 2 toward each other;
at ω t = π / 4 ω t = π / 4 omega t=pi//4\omega t=\pi / 4ωt=π/4 net torque is zero;
at ω t = π / 2 ω t = π / 2 omega t=pi//2\omega t=\pi / 2ωt=π/2 net torque pushes 1 and 2 away from each other.
The angular frequency of the oscillating torque is 2 ω 2 ω 2omega2 \omega2ω. If 2 ω 2 ω 2omega2 \omega2ω is set equal to ω 0 ω 0 omega_(0)-=\omega_{0} \equivω0 (natural oscillation frequency of the arms), the oscillating torque drives the arms into resonant oscillation. The resulting oscillation amplitude, in the 1970 prototype
of the gradiometer, was easily detectable for gravity gradients (Riemann curvatures) of magnitude
0.0002 [ 2 ( mass of earth ) (radius of earth) ] 1 × 10 30 cm 2 .01 g / cm 3 0.0002 2 (  mass of earth  )  (radius of earth)  1 × 10 30 cm 2 .01 g / cm 3 {:[ >= 0.0002[(2(" mass of earth "))/(" (radius of earth) ")]],[qquad1xx10^(-30)cm^(-2)∼.01g//cm^(3)]:}\begin{aligned} & \geq 0.0002\left[\frac{2(\text { mass of earth })}{\text { (radius of earth) }}\right] \\ & \qquad 1 \times 10^{-30} \mathrm{~cm}^{-2} \sim .01 \mathrm{~g} / \mathrm{cm}^{3} \end{aligned}0.0002[2( mass of earth ) (radius of earth) ]1×1030 cm2.01 g/cm3
[ Riem iemn curvature produced by a two-kilometer high mountain, idealized as a two-kilometer high cube, at a distance of 15 kilometers. (Neglected in this idealization are isostacy and any lowering of density of Earth's crust in regions of mountain uplift.) ]  Riem iemn curvature produced by a two-kilometer   high mountain, idealized as a two-kilometer high   cube, at a distance of  15  kilometers. (Neglected   in this idealization are isostacy and any lowering   of density of Earth's crust in regions of mountain   uplift.)  [[" Riem iemn curvature produced by a two-kilometer "],[" high mountain, idealized as a two-kilometer high "],[" cube, at a distance of "15" kilometers. (Neglected "],[" in this idealization are isostacy and any lowering "],[" of density of Earth's crust in regions of mountain "],[" uplift.) "]]\left[\begin{array}{l} \text { Riem iemn curvature produced by a two-kilometer } \\ \text { high mountain, idealized as a two-kilometer high } \\ \text { cube, at a distance of } 15 \text { kilometers. (Neglected } \\ \text { in this idealization are isostacy and any lowering } \\ \text { of density of Earth's crust in regions of mountain } \\ \text { uplift.) } \end{array}\right][ Riem iemn curvature produced by a two-kilometer  high mountain, idealized as a two-kilometer high  cube, at a distance of 15 kilometers. (Neglected  in this idealization are isostacy and any lowering  of density of Earth's crust in regions of mountain  uplift.) ]
For a mathematical analysis of the gradiometer, see exercise 16.5 .
( torque acting on mass 1 relative to center of gradiometer ) i ^ = ϵ i j k k ξ j ( m R k 0 ^ 0 ^ ^ ξ i ) .  torque acting on mass  1  relative to center of   gradiometer  i ^ = ϵ i j k k ξ j m R k 0 ^ 0 ^ ^ ξ i . ([" torque acting on mass "1],[" relative to center of "],[" gradiometer "])_( hat(i))=epsilon_(ijkk)xi_(j)(-mR_({:k( hat(0))( hat(0))( hat(ℓ))xi_(i)).):}\left(\begin{array}{l} \text { torque acting on mass } 1 \\ \text { relative to center of } \\ \text { gradiometer } \end{array}\right)_{\hat{i}}=\epsilon_{i j k k} \xi_{j}\left(-m R_{\left.k \hat{0} \hat{0} \hat{\ell} \xi_{i}\right) .}\right.( torque acting on mass 1 relative to center of  gradiometer )i^=ϵijkkξj(mRk0^0^^ξi).
The torque on mass 4 is identical to this (replace ξ ξ xi\xiξ by ξ ξ -xi-\xiξ ), so the total torque on arm 1-4 is twice this. The components R k ^ 0 ^ o ^ R k ^ 0 ^ o ^ R_( hat(k) hat(0)ℓ hat(o))R_{\hat{k} \hat{0} \ell \hat{o}}Rk^0^o^ of Riemann can be regarded as components of a 3 × 3 3 × 3 3xx33 \times 33×3 symmetric matrix. By appropriate orientation of the reference frame's spatial axes (orientation along "principal axes" of R k k ^ l ^ ) R k k ^ l ^ ) R_(k hat(k)ℓ hat(l)))R_{k \hat{k} \ell \hat{l})}Rkk^l^) ), one can make R k k ^ ^ R k k ^ ^ R_(k hat(k)ℓ hat(ℓ))R_{k \hat{k} \ell \hat{\ell}}Rkk^^ diagonal at some initial moment of time
R x ^ 0 ^ x ^ 0 ^ 0 , R y ^ 0 ^ y ^ 0 ^ 0 , R z ^ 0 ^ z ^ 0 ^ 0 , all others vanish. R x ^ 0 ^ x ^ 0 ^ 0 , R y ^ 0 ^ y ^ 0 ^ 0 , R z ^ 0 ^ z ^ 0 ^ 0 ,  all others vanish.  R_( hat(x) hat(0) hat(x) hat(0))!=0,R_( hat(y) hat(0) hat(y) hat(0))!=0,R_( hat(z) hat(0) hat(z) hat(0))!=0," all others vanish. "R_{\hat{x} \hat{0} \hat{x} \hat{0}} \neq 0, R_{\hat{y} \hat{0} \hat{y} \hat{0}} \neq 0, R_{\hat{z} \hat{0} \hat{z} \hat{0}} \neq 0, \text { all others vanish. }Rx^0^x^0^0,Ry^0^y^0^0,Rz^0^z^0^0, all others vanish. 
Assume that Riemann changes sufficiently slowly along the gradiometer's world line that throughout the experiment R j 0 k ^ 0 ^ R j 0 k ^ 0 ^ R_(j0 hat(k) hat(0))R_{j 0 \hat{k} \hat{0}}Rj0k^0^ remains diagonal and constant. For simplicity, place the gradiometer in the x ^ , y ^ x ^ , y ^ hat(x), hat(y)\hat{x}, \hat{y}x^,y^-plane, so it rotates about the z ^ z ^ hat(z)\hat{z}z^ axis with angular velocity ω 1 2 ω 0 ω 1 2 ω 0 omega~~(1)/(2)omega_(0)\omega \approx \frac{1}{2} \omega_{0}ω12ω0 :
( Angle of arm 1-4 relative to x ^ axis ) = ω t (  Angle of arm 1-4   relative to  x ^  axis  ) = ω t ((" Angle of arm 1-4 ")/(" relative to "( hat(x))" axis "))=omega t\binom{\text { Angle of arm 1-4 }}{\text { relative to } \hat{x} \text { axis }}=\omega t( Angle of arm 1-4  relative to x^ axis )=ωt
Show that the resultant equation of oscillation is
α ¨ + α ˙ / τ 0 + ω 0 2 α = 1 2 ( R x ˙ 0 ^ x ^ 0 ^ R y ^ 0 ^ y ^ 0 ^ ) sin 2 ω t α ¨ + α ˙ / τ 0 + ω 0 2 α = 1 2 R x ˙ 0 ^ x ^ 0 ^ R y ^ 0 ^ y ^ 0 ^ sin 2 ω t alpha^(¨)+alpha^(˙)//tau_(0)+omega_(0)^(2)alpha=(1)/(2)(R_(x^(˙) hat(0) hat(x) hat(0))-R_( hat(y) hat(0) hat(y) hat(0)))sin 2omega t\ddot{\alpha}+\dot{\alpha} / \tau_{0}+\omega_{0}^{2} \alpha=\frac{1}{2}\left(R_{\dot{x} \hat{0} \hat{x} \hat{0}}-R_{\hat{y} \hat{0} \hat{y} \hat{0}}\right) \sin 2 \omega tα¨+α˙/τ0+ω02α=12(Rx˙0^x^0^Ry^0^y^0^)sin2ωt
and that the steady-state oscillations are
α = Im { 1 2 ( R x ^ 0 ^ x ^ 0 ^ R y ^ 0 ^ y ^ 0 ^ ) 2 ω 0 ( ω 0 2 ω + i / 2 τ 0 ) e i 2 ω t } α = Im 1 2 R x ^ 0 ^ x ^ 0 ^ R y ^ 0 ^ y ^ 0 ^ 2 ω 0 ω 0 2 ω + i / 2 τ 0 e i 2 ω t alpha=Im{((1)/(2)(R_( hat(x) hat(0) hat(x) hat(0))-R_( hat(y) hat(0) hat(y) hat(0))))/(2omega_(0)(omega_(0)-2omega+i//2tau_(0)))e^(i2omega t)}\alpha=\operatorname{Im}\left\{\frac{\frac{1}{2}\left(R_{\hat{x} \hat{0} \hat{x} \hat{0}}-R_{\hat{y} \hat{0} \hat{y} \hat{0}}\right)}{2 \omega_{0}\left(\omega_{0}-2 \omega+i / 2 \tau_{0}\right)} e^{i 2 \omega t}\right\}α=Im{12(Rx^0^x^0^Ry^0^y^0^)2ω0(ω02ω+i/2τ0)ei2ωt}
Thus, for fixed ω ω omega\omegaω (e.g., 2 ω = ω 0 2 ω = ω 0 2omega=omega_(0)2 \omega=\omega_{0}2ω=ω0 ), by measuring the amplitude and phase of the oscillations, one can learn the magnitude and sign of R x ^ 0 ^ x ^ 0 ^ R y ^ y ^ y ^ 0 ^ R x ^ 0 ^ x ^ 0 ^ R y ^ y ^ y ^ 0 ^ R_( hat(x) hat(0) hat(x) hat(0))-R_( hat(y) hat(y) hat(y) hat(0))R_{\hat{x} \hat{0} \hat{x} \hat{0}}-R_{\hat{y} \hat{y} \hat{y} \hat{0}}Rx^0^x^0^Ry^y^y^0^. The other differences, R y ^ 0 ^ y ^ O ^ R y ^ 0 ^ y ^ O ^ R_( hat(y) hat(0) hat(y) hat(O))-R_{\hat{y} \hat{0} \hat{y} \hat{O}}-Ry^0^y^O^ R z ^ 0 ^ z ^ 0 ^ R z ^ 0 ^ z ^ 0 ^ R_( hat(z) hat(0) hat(z) hat(0))R_{\hat{z} \hat{0} \hat{z} \hat{0}}Rz^0^z^0^ and R z ^ z ^ z ^ 0 ^ R z ^ 0 ^ x ^ ^ 0 ^ R z ^ z ^ z ^ 0 ^ R z ^ 0 ^ x ^ ^ 0 ^ R_( hat(z) hat(z) hat(z) hat(0))-R_( hat(z) hat(0) hat(hat(x)) hat(0))R_{\hat{z} \hat{z} \hat{z} \hat{0}}-R_{\hat{z} \hat{0} \hat{\hat{x}} \hat{0}}Rz^z^z^0^Rz^0^x^^0^ can be measured by placing the gradiometer's rotation axis along the x ^ x ^ hat(x)\hat{x}x^ and y ^ y ^ hat(y)\hat{y}y^ axes, respectively.]

  1. *Admittedly, this principle is anthropomorphic: twentieth-century physicists like such theories and even find them effective in correlating observational data. Therefore, Nature must like them too!
  2. *If the distance (03) is given arbitrarily, the resulting four-vertex figure will burst out of the plane. Regarded as a tetrahedron in a three-dimensional Euclidean space, it has a volume given by the formula of Niccolo Fontana Tartaglia (1500-1557), generalized today (Blumenthal 1953) to
    ( volume of n -dimensional simplex spanned by ( n + 1 ) points ) = ( ( 1 ) n + 1 2 n ) 1 / 2 1 n ! | 0 1 1 1 1 1 0 ( 01 ) 2 ( 02 ) 2 ( 0 n ) 2 1 ( 10 ) 2 0 ( 12 ) 2 ( 1 n ) 2 ) 2 1 ( n 0 ) 2 ( n 1 ) 2 ( n 2 ) 2 0 |  volume of  n -dimensional   simplex   spanned by  ( n + 1 )  points  = ( 1 ) n + 1 2 n 1 / 2 1 n ! 0 1 1 1 1 1 0 ( 01 ) 2 ( 02 ) 2 ( 0 n ) 2 1 ( 10 ) 2 0 ( 12 ) 2 ( 1 n ) 2 ) 2 1 ( n 0 ) 2 ( n 1 ) 2 ( n 2 ) 2 0 ([" volume of "],[n"-dimensional "],[" simplex "],[" spanned by "],[(n+1)" points "])=(((-1)^(n+1))/(2^(n)))^(1//2)(1)/(n!)|[0,1,1,1,dots,1],[1,0,(01)^(2),(02)^(2),dots,(0n)^(2)],[1,(10)^(2),0,(12)^(2),dots,(1n)^(2)],[cdots,dots,dots,)^(2),dots,cdots],[1,(n0)^(2),(n1)^(2),(n2)^(2),dots,0]|\left(\begin{array}{l} \text { volume of } \\ n \text {-dimensional } \\ \text { simplex } \\ \text { spanned by } \\ (n+1) \text { points } \end{array}\right)=\left(\frac{(-1)^{n+1}}{2^{n}}\right)^{1 / 2} \frac{1}{n!}\left|\begin{array}{cccccc} 0 & 1 & 1 & 1 & \ldots & 1 \\ 1 & 0 & (01)^{2} & (02)^{2} & \ldots & (0 n)^{2} \\ 1 & (10)^{2} & 0 & (12)^{2} & \ldots & (1 n)^{2} \\ \cdots & \ldots & \ldots & )^{2} & \ldots & \cdots \\ 1 & (n 0)^{2} & (n 1)^{2} & (n 2)^{2} & \ldots & 0 \end{array}\right|( volume of n-dimensional  simplex  spanned by (n+1) points )=((1)n+12n)1/21n!|0111110(01)2(02)2(0n)21(10)20(12)2(1n)2)21(n0)2(n1)2(n2)20|
    which reduces for three points to the standard textbook formula of Hero of Alexandria (A.D. 62 to A.D. 150).
    area = { s [ s ( 01 ) ] [ s ( 02 ) ] [ s ( 12 ) ] } 1 / 2 , 2 s = ( 01 ) + ( 02 ) + ( 12 ) ,  area  = { s [ s ( 01 ) ] [ s ( 02 ) ] [ s ( 12 ) ] } 1 / 2 , 2 s = ( 01 ) + ( 02 ) + ( 12 ) , {:[" area "={s[s-(01)][s-(02)][s-(12)]}^(1//2)","],[2s=(01)+(02)+(12)","]:}\begin{aligned} \text { area }= & \{s[s-(01)][s-(02)][s-(12)]\}^{1 / 2}, \\ & 2 s=(01)+(02)+(12), \end{aligned} area ={s[s(01)][s(02)][s(12)]}1/2,2s=(01)+(02)+(12),
    for the area of a triangle. Conversely, if the four points are to remain in two-dimensional Euclidean space, the tetrahedron must collapse to zero volume. This requirement supplies one condition on the one distance (03). It simplifies the discussion of this condition to take (03) small and (102) to be a right triangle, as above. However, the general principle is independent of such approximations, and follows directly from the extended Hero-Tartaglia formula. It is enough in a locally Euclidean or Lorentz space of n n nnn dimensions to have laid down ( n + 1 ) ( n + 1 ) (n+1)(n+1)(n+1) fiducial points 0 , 1 , 2 , , n 0 , 1 , 2 , , n 0,1,2,dots,n0,1,2, \ldots, n0,1,2,,n, and to know the distance of every other point j , k , j , k , j,k,dotsj, k, \ldotsj,k, from these fiducial points, in order to be able to calculate the distance of these points j , k , j , k , j,k,dotsj, k, \ldotsj,k, from one another ("distances between nearby points in terms of coordinates"; metric as distillation of distance data).
  3. qquad\qquad
  4. *For a discussion of quantum-mechanical factor-ordering problems, see, e.g., Merzbacher (1961), pp. 138-39 and 334-35; also Pauli (1934).
  5. *For this information on the response of rubidium clocks to acceleration, we thank H. P. Stratemeyer of General Radio Company, Concord, Massachusetts.
  6. *Based on Marzke and Wheeler (1964).